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Photographic 

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Ce  document  est  filmi  au  taux  da  reduction  indiquA  ci-dassouc 


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Lee  exemplalres  orlglnaux  dont  la  couverture  en 
papier  est  Imprimte  sont  f  limAs  en  commenpant 
par  Ie  premier  plat  et  en  termlnant  solt  par  la 
dernlAre  page  qui  comporte  une  emprelnte 
d'impression  ou  d'illustration,  solt  par  Ie  second 
plat,  salon  Ie  cas.  Tdus  las  autres  exemplalres 
orlglnaux  sont  fllmte  on  commenpant  par  la 
premlAre  page  qui  comporte  une  emprelnte 
d'impression  ou  d'illustration  et  en  termlnant  par 
la  dernlire  pege  qui  comporte  une  telle 
emprelnte. 

Un  des  symboies  suivants  apparattra  sur  la 
dernlAre  image  de  cheque  microfiche,  selon  Ie 
cas:  ie  symbols  — ►  signlfle  "A  SUIVRE",  Ie 
symbols  V  signlfle  "FIN". 

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flimte  A  des  taux  de  reduction  diff brents. 
Lorsque  ie  document  est  trop  grand  pour  Atre 
reprodult  en  un  seul  cllchA,  il  est  filmA  i  partir 
de  Tangle  sup^rieur  gauche,  de  gauche  H  drolte, 
et  de  haut  an  bas,  en  prenant  Ie  nombre 
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1 

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3 

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4 

5 

6 

NEWCOMB'S 

Mathematical  Course. 

— — -♦- — - 

I.    SCHOOL  COURSE. 

Algebra  for  Schools, I|120 

Key  to  Algebra  for  Schools,       .        ...  1I20 

Plane  Oeometry  and  Trigonometry,  with  Tables,  lAO 

The  Essentials  of  Trigonometry,  ....  1.25 

II.    COLLEGE  COURSE. 

Algebra  for  OoUeges, *i  g^ 

Key  to  Algebra  for  OoUeges,      .       .        .    '  .    '  i'.6o 

Elements  of  Oeometry, j  50 

Plane  and  Sphe  -loal  Trigonometry,  with  Tables!  2.00 

Trigonometry  (separate), i.gO 

Tables  (A^orate), 140 

Elements  of  Analjrtio  Geometry      .  1.50 

Oaloolns  (in  preparation)^ 

Astronomy  (Newoomb  and  Holden,)    .                .  2.BO 

The  same,  Briefer  Course,         ....  .1.40 

HEIifRY  HOLT  d  CO.;  Publishers,  New  York. 


NEWOOMB'8  MATHEMATICAL   COURSE. 


ELEMENTS 


GEOMETRY 


BY 


SIMOlSr  I^EWOOMB 

F^ofeasor  of  MathemaUcs,  United  StaUa  Navy 


THIRD    EDITION,  REVISED. 


N"EW    TOEK 

HENRY  HOLT  AND  COMPANY 

1884. 


/ 


» 


Ck>FTRiaHTSD, 
BY 

HENRY  HOLT  &  (DO. 
1881. 


PEEFAOE. 


Is  the  present  work  is  developed  what  is  commonly  known  m  the 
ancient  or  EucUdian  Geometry,  the  ground  covered  being  neariy  the 
same  as  in  the  standard  treatises  of  Euclid,  Legendre,  and  Chauve- 
net.  The  question  of  the  best  form  of  development  is  one  of  such 
interest  at  the  present  time,  among  both  teachers  and  thinkerw,  as  to 
justify  a  statement  of  the  plan  which  has  been  adopted. 

It  being  still  held  in  influential  quarters  that  no  real  improvement 
upon  Euclid  has  been  made  by  the  modems,  a  comparison  with  the 
ancient  model  will  naturally  be  the  first  subject  of  consideration 
The  author  has  followed  this  model  in  its  one  most  distinctive  fea- 
ture, that  of  founding  the  whole  subject  upon  clearly  enunciated  defi- 
mtions  and  axioms,  and  stating  the  steps  of  each  course  of  reasoning 
in  their  completeness.     By  the  common  consent  of  a  large  majority  of 
educators  the  discipline  of  Euclid  is  the  best  for  developing  the  powers 
of  deductive  reasoning.    If  the  work  had  no  other  object  than  that 
of  teaching  geometry,  a  more  rapid  and  cursory  system  might  hftve 
been  followed;  but  where  the  general  training  of  the  powers  of 
thought  and  expression  is,  as  it  should  be,  the  main  object,  it  be- 
comes important  to  guard  the  pupil  against  those  habits  of  loose 
thought  and  incomplete  expression  to  which  he  is  prone.    This  can 
be  best  done  by  teaching  geometry  on  the  time-honored  plan. 

Notwithstanding  this  excellence  of  method,  there  are  several 
points  in  which  the  system  of  Euclid  fails  to  meet  modem  require- 
ments, and  should  therefore  bo  remodeled.  The  most  decided  failure 
18  in  the  treatment  of  angular  magnitude.  We  find  neither  in  Euclid 
""'  "     "'^  "'^  modem  foiiowers  any  recognition  of  angles  equal  to 

98847 


iv 


PREFACE, 


or  exceeding  180°,  or  any  explicit  definition  of  what  Ib  meant  by 
the  Bum  of  two  or  more  angles.  The  additions  to  the  old  systeir  of 
angular  measurement  are  the  following  two :  v 

Firstly.  An  explicit  definition  of  the  angle  which  is  equal  to  the 
sum  of  two  angles. 

Secondly.  The  recognition  of  the  sum  of  two  right  angles  as  itself 
an  angle.  The  term  ''straight  angle"  has  been  adopted  from  the 
Syllalus  of  the  English  Association  for  the  Improvement  of  Geotnetrieal 
Teaching.  Although  not  unobjectionable,  it  seems  to  be  as  good  a 
term  as  our  language  affords.  The  term  geatrecUe  Winkely  used  by 
the  Germans,  is  more  expressive. 

One  of  the  most  perplexing  questions  which  the  author  has  met  in 
the  preparation  of  the  work  is  that  of  distinguishing  the  definitions  of 
plane  figures  as  lines  and  surfaces.  In  our  recent  text-books  it  is  be- 
coming more  and  more  common  to  define  triangles,  circles,  etc.,  as  por- 
tions of  a  plane  surface.  But,  as  soon  as  analytic  geometry  is  reached, 
the  circle  is  considered  as  a  curve  line  and  the  triangle  as  three  straight 
lines,  while,  even  in  elementary  geometry,  these  terms,  in  a  large  ma- 
jority of  cases,  refer  only  to  the  bounding  lines.  It  has  seemed  to  the 
author  that  the  confusion  thus  arising  can  best  be  avoided  by  defining 
plane  figures  neither  as  mere  lines  nor  mere  surfaces,  but  as  things /orm^d 
by  lines;  to  use  the  specific  term  area  when  extent  of  surface  alone  is 
referred  to;  and  to  use  the  words  cireun\ference,  perimeter,  etc, ,  only  in 
the  sense  in  which  they  are  used  in  higher  geometry. 

Other  leading  features  of  the  work,  which  may  be  briefly  pointed  out, 
are  the  following: 

I.  The  addition  of  an  introductory  book  designed  not  only  to  pre- 
sent the  usual  fundamental  axioms  and  definitions,  but  to  practice  the 
student  in  the  aiialysis  of  geometric  relations  by  means  of  the  eye 
before  instructing  him  in  formal  demonstrations.  The  exercises  in 
sections  34  to  84  are  first  attempts  in  this  direction,  to  which  the 
teacher  may  add  at  pleasure  until  he  finds  that  the  pupil  has 
thoroughly  mastered  the  conceptions  necessary  for  subsequent  use. 

n.  The  application  of  the  symmetric  properties  of  figures  in 
demonstrating  the  fundamental  theorem  of  parallels.  This  system 
has  been  adopted  from  the  Germans. 

III.  After  the  second  book,  the  analysis  if  the  problems  of  con> 


PREFACE.  y 

struction,  whereby  the  pupil  is  led  to  discover  the  construction  by 
reasoning. 

IV.  The  division  of  each  demonstration  into  separate  numbered 
steps,  and  the  statement  of  each  conclusion,  where  practicable,  as  a 
relation  between  magnitudes.  It  is  believed  that  this  system  will 
make  it  much  easier  to  carry  the  steps  of  the  demonstration  in  mind. 

Each  step  is,  when  deemed  necessary,  accompanied  by  a  reference 
to  the  previous  proposition  on  which  the  conclusion  is  founded,  not, 
however,  to  encourage  the  too  frequent  habit  of  requiring  the  pupil 
to  memorize  the  numbers,  but  simply  to  enable  him  to  refer  to  the 
proposition.  He  should  always  be  ready,  if  required,  to  cite  the 
proposition,  but  its  number  in  the  book  is  not  of  such  importance 
that  his  memory  need  be  burdened  with  it.  A  reference  has  not 
been  considered  necessary  after  a  few  repetitions. 

V.  The  theorems  for  exercise  have  been  selected  from  native  and 
foreign  works  with  a  view  to  present  those  best  adapted,  either  by 
their  elegance  or  their  applications  in  the  higher  geometry,  to  inter- 
est the  student.  An  attempt  has  been  made  to  arrange  those  of  each 
book  in  the  order  of  their  difficulty. 

VI.  Some  of  the  first  principles  of  conic  sections  have  been  devel- 
oped for  the  purpose  of  enabling  pupils  who  do  not  intend  to  study 
analytic  geometry  to  have  some  knowledge  of  these  curves.  It  is 
believed  that  a  previous  study  of  these  principles  will  be  a  valuable 
preparation  for  the  advanced  treatment  of  conic  sections. 

Vn.  The  most  difficult  subject  to  treat  has  been  that  of  Propor- 
tion. The  ancient  treatment  as  found  in  Euclid  is  perfectly  rigorous, 
but  has  the  great  disadvantages  of  intolerable  prolixity,  unfamiliar 
conceptions,  and  the  non-use  of  numbers.  The  system  common  in 
our  American  works,  of  treating  the  subject  merely  as  the  algebra  of 
fractions,  has  the  advantage  of  ease  and  simplicity.  But,  assuming, 
as  it  does,  that  geometric  magnitudes  can  be  used  as  multipliers  and 
divisors  on  a  system  which  is  not  demonstrated,  even  for  algebraic 
quantities,  it  is  not  only  devoid  of  geometric  rigor,  but  is  not  prop- 
erly geometry  at  all.  The  author  has  essayed  a  middle  course  between 
these  extremes  which  he  submits  to  the  judgment  of  teachers  with 
some  reserve. 

On  the  annienf.  nvflfnm    mnrrni^ii'lpq  <}.rp  r>/%Tnv.««a^  -srSfH -^  ^^ 


vi 


PRBFAOa. 


their  ratios  by  means  of  their  multiples.  For  mstance,  the  magnitude 
A  is  considered  to  have  to  the  magnitude  B  the  ratio  of  2  to  8  when 
8A  =  2B.  This  system  has  the  undeniable  advantage  of  admitting 
commensurable  and  incommensurable  quantities  to  be  treated  on  a 
uniform  plan.  But  it  has  the  disadvantage  of  not  according  with  the 
natural  and  customary  way  of  thini'.ing  of  the  subject.  When  we  say 
that  the  magnitude  A  is  to  B  as  2  to  8,  we  mean  that  if  A  is  repre- 
sented by  the  number  2,  or  is  divided  into  2  parts,  B  will  be  repre* 
sented  by  8  of  those  parts.  The  author  has  considered  it  more  im- 
portant to  base  the  subject  on  natural  and  customary  modes  of 
thought  than  to  adopt  a  system  simple  and  rigorous,  but  not  so  based. 
The  mode  in  which  he  has  endeavored  to  avoid  the  difficulty,  and 
to  render  the  natural  system  as  rigorous  and  nearly  as  simple  as 
the  other,  will  be  s&m  by  an  examination  of  the  chapter  on  Pro- 
portion. 

Ym.  Another  difficult  subject  is  the  fundamental  relations  of 
lines  and  planes  in  space.  In  presenting  it  the  author  has  been  led 
to  follow  more  closely  the  line  of  thought  in  Euclid  than  that  in 
modem  works.  At  the  same  time  he  is  not  fully  satisfied  with  his 
treatment,  and  conceives  that  improvements  are  yet  to  be  made. 

A  collection  of  notes  on  the  fundamental  principles  of  geometry 
upon  which  the  work  has  been  based  will  be  found  in  the  Appendix. 

The  author  believes,  from  some  trials,  that  the  study  of  geometry 
as  here  presented  can  be  advantageously  commenced  at  the  age  of 
twelve  or  thirteen  years.  No  especial  knowledge  of  algebra  is 
required  for  the  first  three  books,  but  a  previous  familiarity  with 
symbolic  notation  will  facilitate  the  study  of  the  second  and  follow- 
ing books,  and  may  be  found  necessary  to  their  advantageous  use. 
From  the  fourth  book  onward  a  knowledge  of  simple  equations  is 
sometimes  presupposed. 


TABLE  OF  OOJ^TEKTS. 


OHAPTiB  ®^°^  ^'    <^ENERAL  NOTIONS. 

I.  The  Primary  Concepts  of  Geometry '^J 

II.  Comparison  of  Geometric  Magnitudes        a 

III.  Of  Symmetry *.*..".*.*.."*.*.'.*!*..'.'.* 15 

IV.  Logical  Elements  of  Geometry !....!..!!!!.! 18 

BOOK  n.    FUNDAMENTAL  PROPERTIES  OP 
RECTILINEAL  FIGURES. 

I.  Relation  of  Angles 2Jj 

n.  Relations  of  Triangles .......... m 

IIL  Parallels  and  Parallelograms ...... . . '. 53 

IV.  Miscellaneous  Properties  of  Polygons. m 

V.  Problems .f V.' .*.*.*."!!*'.' 76 

VI.  Exercises  in  demonstrating  Theorems ............. '. . . . . .  35 

BOOK  IIL    THE  CIRCLE. 

L  General  Properties  of  the  Circle 93 

IL  Inscribed  and  Circumscribed  Figures ina 

IIL  Propertiesof  Two  Circles *. ." jja 

IV.  Problems  relating  to  the  (Jircle ii;* 

Theorems  for  Exercise ."....................    12« 

BOOK  IV.    OP  AREAS. 

L  Areas  of  Rectangles -og 

IL  Areas  of  Plane  Figures .li 

IIL  Problems  in  Areas *.*.".  !?? 

IV.  The  Computation  of  Areas 154 

Theorems  for  Exercise " 1^ 

Numerical  Exercises •..'."...!!..!!!!!!.. 163 

BOOK  V.    THE  PROPORTION  OP  MAGNITUDES. 

L  Ratio  and  Proportion  of  Magnitudes  in  General '  163 

II.  Linear  Proportion * j«» 

IIL  Proportion  of  Areas ..**' ^qa 

IV.  Problems  in  Proportion OQS 

Theorems  for  Exerfiisfi  «.^ 

' 2iv 


viii  CONTENTS, 

BOOK  VI.    REGULAR  POLYGONS  AND  THE  CIRCLE. 

OHATTBR  PAO> 

I.  Properties  of  Inscribed  and  Circumscribed  Regular  Polygons..  214 

II.  Construction  of  Regular  Polygons 210 

III.  Areas  and  Perimeters  of  Regular  Polygons  and  of  the  Circle. .  224 
Exercises 286 

ly.  Maximum  and  Minimum  Figures 287 

Theorems  for  Exercise 248 

BOOK  VII.    OP  LOCI  AND  CONIC  SECTIONS. 

I.  Lines  and  Circles  as  Loci 244 

II.  Limits  of  Certain  Figures 240 

m.  TheEllipse 262 

IV.  The  Hyperbola 268 

V.  The  Parabola 266 

VI.  Representation  of  Varying  Magnitudes  by  Curves 272 

Exercises 274 


GEOMETRY  OF  THREE  DIMENSIONS. 

BOOK  VIII.    OP  LINES  AND  PLANES. 

I.  Relation  of  Lines  to  a  Plane 277 

II.  Relations  of  two  or  more  Planes 208 

m.  Of  Polyhedral  Angles 806 

BOOK  IX.    OP  POLYHEDRONS. 

I.  Of  Prisms  and  Pyramids 816 

n.  The  Five  Regular  Solids 826 

BOOK  X.    OP  CURVED  SURFACES. 

L  The  Sphere 886 

n.  The  Spherical  Triangles  and  Polygons 847 

III.  The  Cylinder 866 

rV.  The  Cone 868 

BOOK  XI.    THE  MEASUREMENT  OF  SOLIDS. 

L  Superficial  Measurement 864 

IL  Volumes  of  Solids 874 

Problems  of  Computation 880 

Thkobems  fob  Exbbcise  in  Gbometby  of  Thbeb  Dimensions.. .  800 


224 
885 
887 
248 


•  844 
.  248 
.  262 
.  268 
.  266 
.  272 
.  274 


SYMBOLS  AOT)   ABBBEVIATIONS    USED    IN 
DEMONSTRATIONS. 


When  a  step  of  a  demonBtration  leads  to  a  relatioii  of 
two  lines  or  other  magnitudes,  tha  relation  is  expressed  by 
symbols. 

=  equals :  states  that  two  magnitudes  are  equal. 

il  parallel  to :  states  that  two  lines  are  parallel. 

^,  perpendicular :  states  that  two  lines  are  perpendiculai' 
to  each  other. 

=  coincides  with,  or  falls  \ipon :  states  that  two  points, 
lines,  surfaces,  or  figures  coincide  with  each  other. 

In  recitation,  the  teacher  may  find  it  advantageous  to 
have  the  student  recit*?  the  reasoninp^  orally,  but  write  th© 
conclusion  of  each  step  on  the  blackboard.  In  this  case 
symbols  or  abbreviations  of  the  more  common  words  will 
shorteu  the  work.  The  following  are  recommended,  though 
others  are  frequently  used: 

Z ,  angh,  /.  line,  • ,  point, 

Ar,  area,  R,  or  90°,  rigJit  angle, 

0,  or  360°,  circumferencG,         S,  or  180°,  slraigkt  angle. 

These  abbreviations  are  not  generally  used  in  the  printed 
booi^  the  author  believing  that  the  full  word,  in  its  usual 
form,  will  make  a  stronger  impression  on  the  mind  of  the 
beginner  than  any  symbolic  representation  of  it. 


BOOK  I. 
GENERAL  NOTIONS. 

CHAPTER     I. 

THE    PRIMARY  'CONCEPTS    OF    GEOMETRY. 


!•  BefmitiGn,  Gteometry  is  the  science  which 
treats  of  magnitude,  position,  and  fonn. 

Def,  A  geometiio  magaitude  is  that  which  has 
extension  in  space,  or  what  is  familiarly  called  size, 

Geometiic  magnitudes  are  of  three  orders :  solids, 
surfaces,  and  lines.  Besides  these,  there  is  a  fourth 
concept — that  of  a  point. 


/da 


Solids. 

Dtf,  A  solid  is  that  which  has  length,  breadth, 
and  thickness.  Iiength,  breadth,  and  thioknesd  are 
called  the  thr^e  dimensions  of  the  solid. 

All  material  bodies  are  solids  because  they  have  these 
three  dimensions  and  no  more.  The  solids  of  geometry  are 
bodies  supposed  to  have  the  size,  form,  and  mobility  of 
material  solids,  but  no  other  properties. 

Surfaces. 

5t.         TlarP  A        mmmmmC^^^       •_      Xl-   _  A  I'll  1  .« 

,^,  ±.^j .  j^  Biiiiav©  ia  mac  whicn  nas  lengtn  and 
breadth,  but  is  not  supposed  to  have  thickness. 

Example  1.     Let  us  conceive  the  solid  AB  to  be  divided 


2 


OENEBAL  NOTIONS. 


I 


into  two  parts  through  CDEF  without  removing  any  part 

of  it,  so  that  the  two  parts  touch 
each  other.  The  diyision  CDEF 
is  then  a  surface.  If  the  surface 
had  any  thickness  it  would  be  a 
part  either  of  the  one  solid  or  of  the 
other.  But  it  is  only  the  bound- 
ary between  them,  and  therefore 

no  part  of  either,  and  therefore  has  no  thickness. 

Ex.  2.    A  sheet  of  paper  is  really  a  solid,  because  it  must 

have  some  thickness.     But  the  surface  on  which  we  write 

does  not  extend  into  the  paper  at  all,  and  so  has  no  thickness. 

it 

Lines. 

4.  Def.  A  line  is  that  which  has  length,  but  is 
not  supposed  to  have  either  breadth  or  thickness. 

If  we  suppose  a  surface  cut  into  two  parts,  touching  each 
other  that  which  divides  them  is  a  line.  It  forms  no  part 
of  either  surface,  and  therefore  can  have  no  breadth. 

Def.  A  straight  line  is  one  which 
has  the  same  direction  throughout  its 
whole  length. 

Def.  A  curve  line  is  one  no  part 
of  which  is  straight. 


A  straight  line. 


A  curve  line. 


The  Point. 

^n.f«  ■^'C"  ^  ^"i"*  **  ^^^^  '^'^"'^  ^  supposed  to  have 
position,  but  neither  length,  breadth,  nor  thickness. 

nthi;  Ti!  r^^T.*  ?'°^  ™*  '"'»  *^»  parts  touching  each 

part  of  either  line,  and  therefore  has  no  length. 

h^^ii,  '"T!*.*"  "«'««'  ha^  ae  three  dimensions,  length, 
breadth,  and  thickness.     But  we  may  make  a  dot  a^d  tWnk 

r„r  "w  ^°""''  ,  '■''^  ""'  P°'"*  ''""^d  ^  tJ'e  centre  of  the 


QEOMETBIO  FTQURBS. 


3 


Generation  of  Magnitudes  by  Motion. 

6.  A  line  may  be  generated  by  the  motion  of  a  point,  as 
when  we  press  the  point  of  a  pencil  on  paper  and  move  it. 

A  surface  may  be  generated  by  the  motion  of  a  line  as  a 
line  is  generated  by  the  motion  of  a  point. 

A  solid  may  be  considered  as  generated  by  the  motion  of  a 
surface.  The  surface,  as  it  moves,  must  be  supposed  to  leave 
a  mark  in  every  point  through  which  it  passes. 

The  Plane. 

7.  Def,  A  plane  is  a  surface  such  that  a  straight 
line  between  any  two  of  its  points  lies  wholly  on  the 
surface. 

A  plane  is  perfectly  flat  and  even,  like  the  surface  of  still 
water,  or  of  a  smooth  floor. 

Geometric  Figures. 

8.  Bef.  A  figure  is  any  definite  combination  of 
points,  lines,  surfaces,  or  solids. 

Def,  A  plane  figure  is  one  which  lies  wholly  in  a 
plane.    It  is  formed  by  points  and  lines. 
Plane  geometry  treats  of  plane  figures. 

Parallel  Lines. 

9.  Def.    Parallel  straight  lines  are  such  as  lie  in 

the  same  plane,  and  never  meet, 

how  far  soever  they  may  be  ex- 
tended m  both  directions.  ParaUelUnes. 

The  Circle. 

10.  Def,  A  circle  is  a  figure  form- 
ed by  a  plane  curve  line,  every  point 
of  which  is  equally  distant  from  a  point 
within  it  called  the  centre. 

The  circumference  of  a  circle  is  the 
line  which  forms  it, 

Def.  An  arc  of  a  circle  is  a  part  of  the  circumfer- 
ence. 


A  circle. 


IT" 


Ml 


OBNBRAL  NOTIONS. 

The  Angle. 
11.  Dtf.    An  angle  is  a  figure  formed  by  two 

straight  lines  extending  out  from 
one  point  in  different  directions. 

Bef,   The  sides  of  an  angle  are 
the  two  lines  which  form  it. 

c        Dtf,    The  vertex  of  an  angle 

An  angle.  is  the  point  where  the  sides  meet. 

Geometrical  Symbols. 

1».  Any  geometric  concept,  whether  apoiTU,  line, 
surface,  solid,  or  angle,  may  be  represented  by  one  or 
more  letters  of  the  alphabet. 

A  point  is  represented  by  a  single  letter  near  it. 

A  line  is  repifesented  by  one  letter,  or  by  two  or 
more  letters  showing  its  course. 

Other  magnitudes  or  figures  are  represented  by  let- 
ters showing  their  outlines. 

Designation  of  angles.  A  particular  angle  in  a  figure  is 
designated  by  three  letters,  as  ABG,  of  which  the  middle  one 
^18  at  the  vertex,  and  one  of  the  other  two  on  each  side. 
The  angle  is  then  read  ABG. 

When  there  is  only  one  angle  formed  at  a  yertex,  it  may 
be  designated  by  a  single  letter  at  the  yertex.  , 


.  »  > 


CHAPTER     II. 

COMPARISON    OF    GEOMETRIC    MAGNITUDES. 


Mode  of  Comparison. 

^\-^^£'  Two  magnitudes  which  can  be  so  applied 
to  each  other  that  each  shall  coincide  with  the  other 
throughout  its  whole  extent  are  said  to  be  IdenUoally 
equal. 


OOMPABIBOS  OF  AlfOLSa.  f, 

™r2  rtot^  *r°  "'?«^*^de8  can  be  SO  divided  into 
parts  that  each  part  of  the  one  is  identically  equal  to 
a  «e^te  part  of  the  other,  they  a^  said  to  be  equa^ 
e^n^™  J^    "'^"^  ^^^"*  *  magnitude  into  two 


B 


0 


The  point  B  bisects  the  line  A  C. 


To  trisect  a  magnitude  means  to  divide  it  into 
three  equal  parts. 

a.."'^^'*u**  ^^^^'-    '''^^  ^°Sl^«  ^^^  and  DJEJIP 
axe  said  to  be  equal  if  the  angle  ^5(7  can  be  taken  up 


Bqnal  angles. 

t Wprf  ^'^  p  *?  1^  ?^^^^  ^^^  ^^  ^'^^^  "tanner  that 
|a  with  the  side  £JIP,  and  the  side  5^  with  the  side 

15.  Unequal  Angles.     If,  on  thus  applying  the 

angles  to  each  other,  the  side  BA 
r  ills  between  the  sides  UJD  and  EF, 
a»  in  the  dotted  Hue,  then  the  angle 
C5^  (which  is  tlie  same  as  FJEJA) 

— ^.„  ^^j  A^oB  buiiii  uie  angle 

FED,  and  the  angle  FEB  is  said 


I 


I!   i 


6 


OBNEBAL  MOTIONS. 


16.  RE3fABK.    The  magDitude  of  an  angle  does  not 
depend  upon  the  length  of 
its  sides,  but  only  upon  their 
direction.  ^ 

When  we  make  the  sides     /  /    / 

^^and  ^C  coincide,  it  is  only     '  '      ^ 

necessary  that  they  shall  coin- 
cide through  the  length  of  the     ««.     - 

dhorfpr  fiirlp  in  n^,!^,.  +«  +«  ^  xi!      J"»ese  four  angles  are  an  equal,  not- 

snorter  sme,  m  order  to  test  the  withstandinsr  the  difference  in  the 

equality  or   inequality  of   the  ^®°8'^  <>' their  sides. 

angles. 

Symbols  of  Comparison. 

Xt.  The  statement  that  any  two  magnitudes  are 
equal  is  expressfed  by  writing  the  sign  =  between  the 
letters  or  words  which  indicate  them. 

The  statement  expressed  by  the  sign  =,  that  two 
magnitudes  are  equal,  is  called  an  equation. 

The  statement  that  one  magnitude  is  greater  or  less 
than  another  is  expressed  by  writing  the  sign  >  or  < 
between  them,  the  opening  of  the  angle  being  toward 
the  greater  magnitude. 
Examples.    The  expression 

A=z  B 

means  that  the  magnitude  A  is  ectual  to  the  magnitude  B, 
The  expression 

A>  B 

means  that  the  magnitude  A  is  greater  than  the  magnitude  B, 
The  expression 

A  <  B 

means  that  the  magnitude  A  is  lesis  than  the  magnitude  B, 

Sum  and  Difference  of  Magnitudes. 

18.  Def.    The  magnitude  formed  by  joining  two 
or  more  magnitudes  together  is  called  their  sum. 


I 


BUM  A2W  DIFFERENCE.  ^ 

The  flum  of  two  or  more  straight  lines  is  the  line  obteined 
by  putting  them  end  to  end  in  the  same  straight  line. 

The  sum  of  two  angles  ABC  and  PQR  is  the  angle  ABR 


Sum  of  anglea 

vZf  n\Tl^n^  ^\''^^  ^^  *^  *^^  «id«  ^^^  «o  that  the 
vertex  (2  shall  fall  on  the  vertex  B,  and  the  sid^  QR  on  the 
opposite  side  of  BC  from  BA, 

Bef,    If  the  angles  ABQ  and  OBR  are  equal,  ea<5h 

me  line  5C  is  said  to  bisect  the  angle  ABU 

When  from  one  magnitude  a  part  equal  to  an- 

Notation  of  Sum  and  Difference. 

the!Si  I^?;r  w*""'  T,^^i*^^^«  i«  expressed  by  writing 
me  sign  +,  ;7^«*5,  between  them.  ^ 

Examples.  -^^^ ^- o 

Angle  ^^C+  angle  CBR  =  angle  ^5i?. 
Line  AB  +  line  ^C  =  line  A  C. 

wrifTi?!  5^^'*^''''!  ^®*'^^^''  ^"^^  magnitudes  is  expressed  by 

Tmtmg  their  symbols  with  the  sign  -,  minus,  between  the^^ 

the  magnitude  taken  away  being  on  the  riffht.'  ' 

l^XAMPLES.     Angle  ABR  -  angle  ABC=  angle  CBR. 

Line  AC-  line  AB  =  line  ^C. 


o 


-D 


8  OENERAX  NOTIONS. 

ELlnds  of  Angles. 

30.  Dqf.    When  a  straight  line  AB  standing  on 

another  straight  line  CD 
makes  the  angles  ABC  and 
ABD  equal,  each  of  these 
angles  is  called  a  right 
angle,  and  the  Une  AB  is 
said  to  be  pexpendioiilar 
Right  angles.  to  the  line  CD. 

21.  Def,    When  the  two  sides  OA  and  OB  of  an 
angle  go  out  in  opposite  direc- 
tions, so  as  to  be  in  the  same    ^^ ^ 

straight  line,  the  angle  is  called  a  <5 

straight  angle.  straight  angle. 

We  may  conceive  a  straight  angle  to  have  its  vertex  at  any 
point  of  a  straight  line. 

A  straight  angle  is  by  definition  the  sum  of  two  right 
angles,  because  the  sum  of  the  two  right  angles  ABC  and 
ABD  is  (1,8)  the  angle  CBDy  which  is  a  straight  angle. 

23.  Def.  An  acute  angle  is  one  which  is  less 
than  a  right  angle. 


Acute  angle. 


Obtuse  angle. 


23.  Def.  An  obtuse  angle  is  one  which  is  greater 
than  a  right  angle. 

Example  of  forming:  Angles  by  Addition. 

24:c  Let  us  take  a  surface  with  a  circular  boundary 
ABCDEFGH,  and  cut  it  into  eight  equal  parts  by  four  lines 
all  passing  through  its  centre  0. 

A  circular  disk  of  paper  or  pasteboard  may  be  used  to  represent  this 
surface.  * 


ADDITION  OF  ANGLES. 

Then  let  us  put  the  pieces  to- 
gether again,  one  by  one,  beginning 
at  A  and  going  round  in  alpha- 
betical order,  and  let  us  study  the 
angles  thus  formed. 

On  adding  the  angle  once  we 
shall  have  a  right  angle  AOC, 

On  adding  it  again  we  shall 
have  the  obtuse  angle  A  OD,  greater 
than  a  right  angle,  but  less  than  a 
straight  angle. 

On  adding  it  again  we  shall 
have  a  straight  angle  A  OB,  be- 
cause we  cut  the  figure  so  that 
A  OB  should  be  in  a  straight  line. 

On    adding  it  again  we  shall 
ihave  the  convex  angle  A  OF.   This 
angle  will  be  greater  than  a  straight 
angle  if  we  count  it  round  in  the 
direction  we  have  added  its  parts, 
but  it  will  be  less  if  we  count  it  in 
the    shortest    direction    from    A 
through    II  and   G    to   F.     The 
relation  of  these  two  ways  of  con- 
sidering an  angle  will  be  shown 
presently. 

By  one  more  addition  the  angle 
formed  will  be  the  sum  of  a  right 
angle  and  a  straight  angle,  or  of 
three  right  angles,  when  measured 
the  one  way,  but  equal  to  a  right 
angle  when  measured  the  other 
way. 

If  we  add  the  angle  twice  more   -o 
the  whole  space  around  O  will  be 
filled  up,  and  the  sum  of  the  eight 
— £,xvij   TTiii  yu    inQ    anp*' ^  AOA 
counted  all  the  wayrounu,  »fhich  is 
called  a  perigon. 


0 


li 


ll 


I 


10 


GENERAL  NOTIONS. 


Dtf.  A  perigon  is  equal  to  the  sum  of  four  right 
angles  or  of  two  straight  angles. 

Angular  Measure. 

The  following  summary  includes  a  recapitulation  of  results 
from  the  preceding  sections: 

35.  An  angle  is  measured  by  how  much  one  of  the 
sides  must  be  turned  to  make  it  coincide  with  the  other 
side. 

Since  one  side  can  be  brought  into  coincidence  with 
the  other  by  turning  it  in  either  direction,  there  are 
two  measures  to  every  angle. 

Example.  In  the  figures  the  side  OA  can  be  brought  into 
coincidence  with  OD  by  turning  it  either  in  the  opposite 
direction  to  that  in  whtch  the  hands  of  a  watch  move,  or  in 
the  same  direction. 


The  lesser  measure  of  the 
angle  AOD. 


The  greater  measure  of  the 
angle  AOD. 


These  two  directions  can  be  distinguished  and  the  amount 
of  motion  be  measured  by  describing  an  arc  of  a  circle  from 
one  side  to  the  other  around  the  vertex  of  the  angle  as  a 
centre.  This  arc  must  pass  through  the  space  over  which  the 
one  arm  must  turn  in  order  to  coincide  with  the  other. 

36.  In  practice,  angles  are  measured  by  degrees  and  sub- 
divisions of  a  degree,  in  the  following  way: 

Let  a  complete  circle  be  drawn  with  its  centre  on  the 
vertex  of  the  angle.  Let  this  circle  be  divided  into  360  equal 
parts.     Then  each  of  these  parts  is  called  a  degree. 

The  sides  of  the  angle  will  cut  the  circle  in  two  points. 

Thft  Tmrnhfir  nf  HAorrppa  Tip+xirooTi   +1^000  •K»i>i«+r.  ;«  +1,^ , 

of  the  angle,  and  the  angle  is  said  to  be  of  that  number  of 
degrees. 


ANGULAR  MEAiiUME, 


11 


luuuSuru 


The  two  sides  of  the  angle  divide  the  circle  into  two  arcs 
corresponding  to  the  two  measures  of  the  angle  just  de- 
scribed. 

Example.  In  the  figure  the  side  OA  of  the  angle  A  OB 
cuts  the  circle  at  20°,  and  the  side  OB  at  360°  Counting 
the  degrees  in  both  directions  we  see  that  the  angle  measures 
240  m  one  direction,  and  120°  in  the  other. 

oaf^^""^^^'.  P®  i"^  ^*  *^®  *^«  measures  wiU  always  be 
obO  ,  which  IS  therefore  a  perigon. 

27.  Def.  The  two  measures  of  an  angle  are  said 
to  be  conjugate  to  each  other,  or  to  represent  conJu- 
gate  angles.  ^ 

.1,  ^^l^o  *^®  ^'^njiigate  measures  will  always  be  less 
than  180  and  the  other  greater,  except  when  each  is 
equal  to  180°. 

The  greater  measure  is  called  a  reflex  angle.     Hence, 

.iJ^t:  ^  r^^""  ^""^^^  ^®  """^^  ^^i^^  is  greater  than  a 
straight  angle,  or  greater  than  180°. 

been  L?"  ^"^^"'"'''^  ''^^*'°^'  ^^  ^^«^'«  ^^^  ^^"^  ^l^^t  has 


1  perigon 


1  straight  angle 
•  2  straight  angles 


1  right  angle 

2  right  angles 
4  right  angles 


:  90° 
180° 
360° 


pi 


18 


QBNERAL  N0TI0N8. 


BXBRGISBS. 


Note.  The  following  exercises  are  intended  to  familiarize  the  pupil 
with  the  idea  of  the  magnitudes  and  measures  of  angles  by  causing  him 
to  make  an  eye-estimate  of  the  magnitude  of  each  angle,  and,  where 
applicable,  a  computation  of  their  relations.  It  will  be  well  for  him  to 
make  a  small  paper  protractor  in  order  that  he  may  check  his  estimates 
by  some  kind  of  measures,  though  rude. 

Whore  he  is  asked  to  draw  angles,  it  is  Intended  that  he  shall  prac- 
tice the  drawing  without  instruments,  repeating  hi8  first  attempts  until 
he  obtains  a  drawing  as  accurate  as  he  can  make  it>by  the  unaided  eye. 

1.  What  kind  of  an  angle  is  each  of  the  following,  arsJ 
how  many  degrees  do  you  judge  it  measures,  the  magnitude 


a 


-AO 


B- 


o 


-A   "Oi 


of  each  angle  heing  measured  from  OA  to  OB  in  a  direction 
the  opposite  of  that  of  the  motion  of  the  hands  of  a  watch  ? 

2.  What  is  the  magnitude  of  each  of  th^*  ^  ''  n^'ag  angles, 
^0(7and(70J?? 


A      B- 


-A 


B- 


COMPARISON  OF  FIQUliSS. 


18 


3    Draw  an  acute  angle  ^OA     Bisect  it.     Draw  another, 
and  trisect  it.  ' , 


■— A 


trisect  u'^*''  *""  ''^^'''^  ''''^'''"     ^'^''''^  ''^'     ^'^"^  *''^*'^«^  "^^ 

6.  Draw  an  angle  of  176°  and  bisect  it. 

trisect  u''*''  ^  '^'^'^^^  ''''^^'  ^"""^  ^''"'^  '^'    ^'^^  ^"^"^^^  »»d 

7.  Draw  a  reflex  angle  and 
bisect  it  on  the  convex  side.  Then 
bisect  the  conjugate  angle  on  the 
other  side.  Estimate  the  number 
of  degrees  in  each  of  the  angles 
thus  formed. 

8.  Here  are  seven  straight 
lines  going  out  from  the  same 
point  and  making  equal  angles 
with  each  other.  Now  draw  five 
other  figures  formed  respectively 
of  6,  5,  4,  3,  and  2  straight  lines 
going  out  from  the  same  point 
and  making  equal  angles  with 
each  other.  How  many  degrees 
m  each  angle  thus  formed  ? 

9.  Draw,  by  the  eye,  angles  of 
60°,  90°,  120°,  160°,  210°,  240°, 
270°,  300°,  330°. 

Comparison  of  Geometric  Figures. 

89.  The  only  way  in  which  we  can  decide  whether 
two  magnitudes  are  equal  or  unequal  is  by  applying 


n 


I  u 


'  ■  ( 


.( 


Ilillillili 


Mil 


14 


GENERAL  NOTIONS. 


one  to  the  other,  or  applying  some  third  magnitude 
to  both. 

We  are  to  think  of  the  geometric  figures  as  mBdo  of  per- 
fectly stiff  lines  which  can  be  picked  up  from  the  paper  and 
moved  about  without  bo:iding  or  undergoing  any  change  of 
lorm  or  magnitude. 

If  two  straight  lines  are  to  be  compared  we  ^ay  one  upon 
the  other,  and  find  whether  the  two  c  ds  can  be  made  to 
coincide.  If  so  they  are  equal ;  if  not,  unequal.  "We  may 
also  take  some  measure  (a  scale  of  equal  parts,  for  example) 
and  apply  it  first  to  one  line  and  then  to  the  other. 

If  two  planes  are  to  be  compared,  they  may  be  applied 
without  change  to  each  other.  If  they  are  of  different  shapes, 
one  may  be  cut  to  pieces  and  the  parts  laid  upon  the  other. 
If  the  latter  can  thus  be  exactly  covered,  the  two  surfaces  are 
equal ;  if  more  than  covered,  the  first  is  the  larger ;  if  not 
covered,  the  second  is  the  larger. 

Solids  are  compared  by  finding  whether  they  will  fill  the 
same  space,  one  or  both  of  them  being  cut  to  pieces  if  necessary. 

But  the  geometer  does  not  actually  apply  his  figures  to 
each  other,  but  only  imagines  them  so  applied.  He  is  thus 
able  to  learn  things  which  are  true  of  all  figures  of  a  certain 
kind,  whereas  by  actual  measurement  one  can  only  learn  what 
is  true  of  the  particular  figure  which  he  measures.  This  will 
be  better  understood  when  it  is  seen  how  theorems  are  demon- 
strated. 

Trace  of  a  Fig^ure. 

30.  When  we  imagine  a  figure  moved  away,  we  may  also 
imagine  that  it  leaves  its  outline  fixed  upon  the  paper.  Such 
an  outline  is  called  a  trace.  The  trace  will  occupy  exactly  the 
position  which  the  figure  itself  occupied  before  being  moved, 
will  be  equal  to  the  figure  in  every  respect,  and  will  be  repre- 
sented by  the  drawing  of  the  figure.  If  another  figure  is 
found  to  coincide  with  the  trace,  it  will  be  identically  equal 
to  the  first  figure. 

Since  figures  are  supposed  to  be  movable,  the  beginner 
may  grasp  the  relation  between  a  figure  and  its  trace  by 
imagining  that  he  marks  around  the  figure  with  a  pencil. 
Then  when  the  figure  is  taken  away  the  marks  will  remain. 


aTMMETBF. 


15 


CHAPTER     III. 

OF  SYMMETRY. 


Symmetry  with  Respect  to  an  Axis. 

31.  Let  us  take  the  figure  in  the  murgin,  turn  it  over, 

and  put  It  back  so  that  the  line  PQ 
shall  fall  in  its  original  position,  but 
the  figure  shall  bo  turned  right  side 
left.     It  will  then  fall  into  the  position 
^    represented  by  the  dotted  lines      "J'o 
make  this  change  of  jiosition  we  may 
suppose  the  ends  of  the  line  PQ  io  be 
pivots,  so  that  the  figure  can  turn  on 
this  line  as  on  an  axis.     In  turning 
one  side  of  the  figure  must  be  sup' 
An  unsymraetricai  figure.    P^^^^^  ^^  sink  below  the  paT)er  and  the 

^Gremair.unmove.l.""""'  '"  "^  "'"''"  '*'  ^^"«  '''^  "^ 

If  we  take  this  figure  and   turn  it 
over  on  the  axis  PQ,  the  right  side  will 
tall  on  the  trace  of  the  left  side,  and 
"««  versa,  so  that  the  figure  will  oo 
cupy  the  same  lines  on  the  paper  as  be- 
fore It  was  moved.     Such  a  flgare  is 
said  to  be  symmetrical  with  respect  to 

nWon-"  ^'"""'  ""^  ^"^^""^'k defi- 

»«f  ^'  J^f-  .^  *»"'*  '»  said  to  be  Q 

tw  '       ^^S  turned  over  on    ""^  *"  «"<'  "^'^  ^• 

I'lT,*^?"'  every  part  of  the  fiimre  i«  ,>  ti-g  ..«-^«- 

symmetry  "-evolution  is  called  an  axis  of 


! 


,  1, 


1 


\Vl 


f!  n 


It 


J  11  I!  ; 


it ' 


I  I 


16 


GENERAL  NOTIONS. 


BXBRCISBS. 


1.  Copy  the  following  figures.  Then  imagine  each  one 
turned  over  so  that  the  line  ^i^  shall  be  changed  end  for  end, 
and  draw  dotted  lines  showing  where  the  rest  of  the  figure 
would  fall. 


E- 


— ^r  E' 


— F  E- 


-FE 


/ 


E^ — ^'  J 


E- 


-F    E- 


-F    E 


-F 


2.  Let  each  of  the  following  figures  be  turned  over  on  the 
line  PQ  us  an  axis.  Then  draw  dotted  lines  showing  where 
the  figure  will  fall. 


3.  Draw  the  axis  of  symmetry  of  each  of  the  following 
figures.    If  there  is  more  than  one  such  axis,  draw  them  all. 


4.  How 


many  axes  of  symmetry  can  be  drawn  to  a  circle  ? 


aYMMETRT. 


17 


Symmetry  with  Respect  to  a  Point. 


,33.  We  may  next  suppose  that  the  figure,  instead  of 
being  turned  over  IS  turned  half  way  round  on  a  fixed  point 
without  leaving  the  paper.     For  instance,  suppose  the  an 


A  flgupe  unsymmetrloal  with  respect  (o  the  potat  11. 

neied  figure  to  have  a  pin  stuck  through  it  at  the  point  M 
and  to  be  umed  half  round  on  that  jln.  It  will  then  tal« 
up  the  position  shown  by  the  dotted  outline.  ' 

„,i  T!™n ''''  ''«"'*  •"*"  ^"y  "^""""J  in  *e  same  way,  everv 
part  of  It  wiU  occupy  the  position  ^'     ^ 

which  the  opposite  part  occupied 
before  the  motion,  and  the  position 
of  the  figure  will  be  represented  by 
the  same  drawing.  Such  a  figure  is 
said  to  be  symmetrical  with  respect 
to  the  point  M. 

Hence  the  following  definition: 

34.  D^.  A  figure  is  said  to        .<~-^». «.  ««>  point « 
h?rn!?r  u'**''^  "^^  ""P*"*  *»  »  point  when,  being 

S  is  SL'°r''  "\^j^  p°^"*'  «-ry  P-'  of  th« 


A  figure  symmetrical  with 
respect  to  the  point  M. 


to  a  circle  ? 


I  h 


iiiij 


1 


18 


GENERAL  NOTIONS. 


In  the  latter  the  figure  does  not  leave  the  paper,  but 
simply  turns  on  it  without  turning  oyer.  Every  part  of  the 
figure  changes  places  with  the  part  which  is  at  an  equal  dis- 
tance on  the  other  side  of  the  pivot  point. 


EXERCISES. 


Copy  the  following  figures.  Then  suppose  them  turned 
half  way  round  on  the  point  M,  and  draw  dotted  lines  show- 
ing where  the  figure  will  fall. 


M 


M 


M 


M 


^/ZA 


> » > 


CHAPTER    IV, 
LOGICAL  ELEMENTS  OF  GEOMETRY. 


Definitions. 

35-  Def.  A  proposition  is  either  a  statement  that 
something  is  true,  or  a  requirement  that  somethiug 
shall  be  done. 

A  proposition  affirming  something  to  be  true  may 
be  either  an  axiom  or  a  theorem. 

36.  Def.  An  axiom  is  a  statement  which  we  as- 
sume to  he  true  without  proof. 

For  the  axioms  of  geometry  we  try  tc  ^ake  propositions 
which  are  self-evident  and  so  need  no  proof. 


AXIOMS  OF  OEOMETRT. 


19 


37.  Def.  A  theorem  is  a  statement  which  requires 
to  be  proved. 

A  proposition  requii^ng  something  to  be  d(Jne  may 
be  a  postulate  or  a  problem. 

38.  Def.  A  postulate  is  something  which  we  sup- 
pose capable  of  being  done  without  showing  how. 

39.  Bef,  A  problem  is  something  which  we  must 
show  how  to  do. 

40.  Def.  A  demonstration  is  the  course  of  rea- 
soning by  which  we  prove  a  theorem  to  be  true. 

41.  Def.  A  corollary  is  a  theorem  which  follows 
from  some  other  theorem. 

43.  Def.  A  lemma  is  an  auxiliary  theorem,  to  be 
used  in  demonstrating  some  other  theorem. 

43.  Def.  A  soholium  consists  of  remarks  upon 
the  application  of  theorems. 

Axioms  of  Geometry. 

44.  Axioms  of  magnitude  in  general. 

Axiom  1.  Magnitudes  which  are  each  equal  to  the 
same  magnitude  are  equal  to  each  other. 
Symbolic  expression  of  this  axiom.     From 

Magnitude  X  =  magnitude  A, 
Magnitude  Y  =  that  same  magnitude  A, 
we  conclude 

Magnitude  X=  magnitude  Y. 

Ax.  2.  If  equals  bemadded  to  equals,  the  sum  will 
be  equal. 

Symbolic  expression.    From 


we  conclude 
and 


X=  Y, 
A  =  B, 


T  -^ 


V .    r> 


jj 


■XT 


r  J- 


A-\-Y=B^X. 


j  1 

;  i 

1 
i   ^ 

1    t ;  ■      i 

i  1 
i 

1       : 

!                 :       i 

1 

iii  }' 

ii  i ' 
it  l! 


I!!  I 


20 


OBNBRAL  NOTIONS. 


Ax*  3.  If  equals  be  subtracted  from  equals,  the 
remainders  will  be  equal. 

Symbolic  expression.    From     ^  ^ 


we  conclude 


A  z=  B, 
X-A==r-B. 


Ax.  4.  Similar  multiphjs  of  equals  are  equal  to 
each  other. 

Symbolic  expression.    If  n  be  any  number,  then  from 

X==A 
we  conclude 

n  times  JT  ==  /i  times  A, 
This  may  be  regarded  as  a  corollary  from  Axiom  2. 

Ax.  5.  Similar  fractions  of  equal  magnitudes  are 
equal. 

Ax.  6.  If  equals  be  added  to  unequals,  that  sum 
will  be  the  greater  which  bias  been  obtained  from  the 
greater  magnitude. 

Symbolic  expression.    From 


we  conclude 


A  >  B, 

A-{-x>  B^  r. 


Ax.  7.  If  equals  be  subtracted  from  unequals,  that 
remainder  will  be  the  greater  which  is  obtained  from 
tne  greater  minuend. 

Symbolic  expression.    From 


we  conclude 


A  >  B, 
X=Y, 


A-  X>  B-  r. 
Ax.  8.  The  whole  is  greater  than  its  part. 


DEMOIfSTBATION  OF  THEOREMS.  gj 

45.  Axioms  of  geometric  relation. 
Ax  9.  A  straight  line  is  the  shortest  distance  bfi 
tween  any  two  of  its  points.  ^isrance  be- 

Ax.10.  If  two  straight  lines  coincide  in  two  or 
moy  mts,  they  will  coincide  throughout  their  whole 

asiSS.  ^^^^^^^^^----tersectinonly 

Ax.  11.  Through  a  given  point  one  straight  lin« 
can  be  dmwn  and  only  one,  which  shall  be  Mel  to 
a  given  straight  line.  P^^ranei  to 

The  Demonstration  of  Theorems. 

[encJto  a'C^"  "  ^^^'""^  *»  demonstration  by  refer- 
the'hj^S*'on^hrot^  'r".  "f'  "»'  eo^espondto 
fnlfiUingthecondiS  '  *™'^  P°^'"«  ««"«> 

designate  ttom,  mZv  a«,  ™^^  ,^1"'  ^°'  '""'^  ""'y  "^  ««««  to 
(reader.  •  "  ™y  «««  supposed  to  be  conceiyed  in  tlie  mind  of  tlio 

thaf'ti./vil^^?  *^°  propositions  are  so  relatfi,? 


22  GENERAL  HOTIONS. 


il 


m 


nilLhi      .:! 


li lit  ill 
liiiiiii  "■ 


Theorem  I.* 

48,  A  straight  line  can  he  bisected  in  only  a 
single  point. 

Here  the  hypothesis  supposes  that  we  take  any  straight 
line  whatever  and  bisect  it.  p 

To  enunciate  the  hypothesis  we  call    a  o  b 

one  end  of  the  line  A  and  the  other  end  B,  and  the  point  of 
bisection  0. 

Then  the  hypothesis  means  that  the  point  0  is  equally 
distant  from  A  and  B. 

The  conclusion  asserts  that  there  is  no  other  point  than  0 
on  the  line  which  is  equally  distant  from  A  and  B. 

The  proof  is  ^fEected  by  showing  that  to  suppose  any 
other  point  having  this  property  is  impossible. 

If  there  is  such  a  point,  call  it  Py  and  suppose  it  between 
A  and  0  (because  we  may  call  either  end  of  the  line  A), 

Let  us  then  suppose  that  PA  is  equal  to  PB. 

Because  P  is  between  A  and  0,  AP  will  be  less  than  A  0, 

Because  OB  is  by  hypothesis  equal  to  OA,  PB,  which  is 
greater  than  OB,  will  be  greater  than  OA. 

Therefore,  if  we  suppose  PA  and  PB  equal,  PA  will 
be  greater  than  OA  and  less  than  OA  at  the  same  time, 
which  is  absurd.  Therefore  there  is  no  point  on  the  line 
except  0  which  is  equally  distant  from  the  ends  of  the  line. 

Theorem  II. 

49.  ^  straight  line  is  symmetrical  with  respect 
to  the  perpendicular  passing  through  its  middle 
point 

Hypothesis.  AB,a  straight  line;  0,  its  middle  point;  PQ, 
a  perpendicular  passing  through  0. 

*  These  simple  theorems  are  presented  partly  as  exercises  and  explana- 
tions for  the  beginner,  and  partly  as  the-  basis  of  subsequent  theorems. 
The  demonstrations  are  not  necessarily  to  be  recited  in  full  as  given, 
but  the  student  should  be  encouraged  and  assisted  in  stating  the  sub- 
stance of  the  reasoning  in  his  own  language. 


DEMONSTRATION  OF  THEOREMS. 


23 


appose  any 


-B 


Conclusion,     The  line  AB  is  symmetrical  with  respect  to 
the  axis  PQ. 

By  reference  to  the  definition  of  symmetry,  §  83,  the  conclusion 

is  found  to  mean  that  if  the 
P  line  AB  be  turned  over  on  the 

line  P^  as  an  axis,  it  will  fall 
on  its  own  trace;  that  is,  into  its 
original  position;  being  merely 
changed  end  for  end. 

Demonstration.  Sup- 
pose the  line  turned  over 
on  the  axis  PQ,  By  hy- 
pothesis and  definition  the 
angles  FOB  and  POA  are 
^  equal.   Therefore,  after  the 

nA     -n  *  11  •  .     X.  ^*"®  ^^  turned  over,  the  side 

OAmW  fall  into  the  position  OB,  and  vice  versa.     (8  14) 

Because  the  lengths  OA  and  OB  are  equal  (by  hypothesis  ] 
the  point  ^  will  fall  on  5,  and  vtce  versa. 

So  the  conclusion  is  proved. 

Exercise  for  the  pupil  Prove  in  the  same  way  that  the 
line  AB  is  symmetrical  with  respect  to  the  point  0  as  a  centre 
of  symmetry  (§  34). 

Theorem  III. 

50.  All  straight  angles  arid  all  right  angles  are 
equal  to  each  other,  y      ^/c 

To  prove  the  first  part  of  this  proposition  it  is  sufficient' 
to  show    that    any  two 

straight  angles  we  choose    A— Q g 

to  take  are  equal. 

The    hypothesis   will  ^ 

be  that  we  have  any  two   M ^         __j^ 

straight  angles  which  we  may  call  A  OB  and  MQN. 

By  the  definition  of  a  straight  angle  the  hypothesis  will 
mean  that  04  and  OB  go  out  from  0  in  opposite  directions 

nM     /^?. ''  *  '*^^^^^*  ^^^®'  »^<i  *^at  thi©  same  is  true  of 
^^and  OJV^. 

mu^XavJ^  ?^*?'°^  ""^  ^^'^  hypothesis  apart  from  its  enunciation 
must  always  be  clearly  apprehended  by  the  student. 


84 


GENERAL  NOTIONS. 


m 


The  conclusion  is  that  the  angles  A  OB  and  MQN  are 
equal. 

By  the  definition  of  equality  in  angles,  (§  14),  this  will 
mean  that  if  we  apply  A  OB  to  MQN  so  that  the  side  A  0  shall 
coincide  with  MQy  then  the  side  OB  will  coincide  with  QN, 

^  This  must  be  the  case, 

^ — — — ^B   because  two  straight  lines 

coincide  throughout  when 
Q  any  two  of  their  points 


— N   coincide.    (§  46,  Ax.  10). 


M 

Therefore  the  conclusion  is  proved,  because,  from  the  fact 
that  any  two  straight  angles  are  equal,  it  follows  that  all  are 
equal. 

Because  a  right  angle  ia,  by  definition,  the  half  of  a 
straight  angle,  and  because  all  straight  angles  are  equal,  it 
follows  from  §  44,  Axiom  6,  that  all  right  angles  are  equal. 

Theoeem  IV. 

51.  The  sum  of  all  the  angles  formed  on  one  side 
of  a  straight  line  by  lines 
emanating  from  a  point  mi 
it  is  a  straight  angle. 

Proof.  If  0  be  the  point  from 
which  the  lines  emanate  and  OB, 
OC,  etc.,  the  lines  emanating  ^ 
from  it,  then,  by  definition  (§  18),  the  sum  of  all  the  angles, 
AOB,  BOO,  etc.,  to  DOE,  will  be  the  angle  ^(9J?;  that  is, 
a  straight  angle;  because  ^O^is  a  straight  line. 

62.  Corollary.  The  sum  of  all  the  angles  around  a  point 
is  equal  to  two  straight  angles,  or  a  circumference. 


ill! 


Itlll! 


BOOK  11. 

FUNDAMENTAL   PROPERTIES    OF 
RECTILINEAL  FIGURES, 


CHAPTER    I. 
REUTIONS  OF  ANGLES. 


Definitions. 


53.  Def.  A  rectilineal  figure  is  one  which  is 
fonned  by  straight  lines. 

54.  Def.  A  triangle  is  a  figure  formed  by  three 
straight  lines  joined  end  to  end. 

55.  Def.  The  three  lines 
which  form  a  triangle  are  called 
its  sides. 

56.  The  sides  of  a  triangle 
may  be  produced  indefinitely.  a  triangle. 
It  is  then  called  a  general  triangle. 

Remark.  Any  three  indefinite  straight  lines  which  inter- 
sect each  other  in  three  different  points  form  a  general  tri- 
angle.   See  figure  on  following  page, 

57.  An  interior 
angle  of  a  triangle  is 
one  between  two  sides, 
measured  inside  the 
triangle. 

58.  An    exterior, 


angle  of  a  triangle  is 


Bbrterior  angle. 


nrio  w 


lllrt"K     ■Jo     fr\^im\r\A   1~..^4-'..r.^^..    ^_.^    _J  J J    xT xJ 

xa  xviiiicu.  MCl/WCCli  a-uy  JSIUfJciliU.  Liie  coiiiinu* 


ation  of  another  side. 


i        HI 


,  I 


■I!: 
1)1  in 


96 


BOOK  11.    RECTILINEAL  FIGUREB. 


NoTB.    The  general  triangle  has  six  exterior  angles  in  all. 

Remark.    When  no  adjective  is  applied  to  the  angle  of  a 
triangle  an  interior  angle  is  meant.  ^ 


The  six  exterior  angles,  e  c  e  e  e  «,  of  the  general  triangle. 

69.  Def.  When  the  sum  of  two  angles  is  a  right 
angle,  each  is  said  to  be  the  oomplement  of  the 
other,  and  the  two  are  called  oomplementary  angles. 


Complementary  angles.  ABC  and  CBD  are  supplementary  angles. 

60.  Def.  When  the  sum  of  two  angles  is  a 
straight  angle,  each  is  said  to  be  the  supplement  of  the 
other,  and  the  two  are  called  supplementary  angles. 

By  definition,  the  sum  of  two  angles  ABG  and  CBD  will 
be  a  straight  angle  when  the  two 
sides  AB  and  BD  lie  in  the  same 
straight  line.    Therefore:  0<j? 

Corollary.  If  two  supplemen- 
tary angles  he  added,  their  extreme 
sides  will  form  a  straight  line.  «  »«<» ».  *  and  c,  c  and  d,  d  and 

<*  »"  adjacent  angles ;  o  and  c  are 
bl.   When    two    straight    lines   **PP<*^**»n«les,and8oare6andd. 

cross  each  other  four  angles  are  formed,  which  we  mav  call 
a,  o.  Cy  and  d. 


RELATIONS  OF  ANGLES. 


97 


Any  two  of  those  angles  which  adjoin  each  other,  as  a  and 
h,  are  called  adjaoent  anglei. 

By  §  60  two  such  adjacent  angles  are  supplementary. 

62.  Drf.  A  pair  of  angles  contained  between  the 
same  two  lines  on  opposite  sides  of  the  vertex  are 
called  opposite  angles. 

63.  Def.  A  transversal  is  a  straight  line  cross- 
ing several  other  lines. 

64.  Angles  formed  by  a  transversal  When  a  transversal 
crosses  two  parallel  lines  it 
forms  with  them  eight  angles, 
four  on  each  line.  Pairs  of 
these  angles  are  designated 
thus : 

The  angles  a,  ft,  g,  and  h  are 
called  eiEterior  angles.  The 
angles  c,  d,  e,  and /are  called 
interior  angles.  The  pair  c 
and  e  and  the  pair  d  and/,  on  opposite  sides  of  the  transversal 
and  between  the  parallels,  are  called  alternate  angles. 
The  pairs  a  and  e,  b  and  /,  c,  and  g,  d  and  A,  on  correspond- 
ing sides  of  the  parallels  and  transversal,  are  called  corre- 
sponding angles. 

Remarks  on  Straigrht  Lines. 

Q5.  Every  straight  line  may  extend  without  end 
in  both  directions.  It  is  then  called  an 
indefinite  straight  Une. 

Sometimes  we  have  to  consider  a  line  extend- 
— g  ing  out  only  in  one  direction,  and  terminating 
at  a  point  in  the  other  direction. 

Example.     In  considering  an  angle,  the  two 
sides  are  suppose  to  terminate  at  the  vertex. 

Sometimes  we  have  to  consider  a  straight  line  con- 
tained between  two  definite  points  which  are  its  ends. 
Such  a  piece  is  called  a  finite  straight  line. 


28 


BOOK  II.    RECTILINEAL  FIGUHES. 


Mm 


66.  JOf.  The  perpendicular  bisector  of  a  finite 
straight  me  IS  an  indefinite  stmightUne  passing  a^ 
right  angles  through  its  middle  point. 

Theorem  I. 

67.  If  two  straight  lines  intersect  each  other,  the 

opposite  angles  will  be 
equal. 

Hypothesis.  AB  and 
CD,  two  straight  lines;  0, 
their  point  of  intersec- 
tion; a,  the  angle  DOB; 

,    ^^  ,  «'» tlie  angle  AOC;  h,  the 

angle  DO  A;  h',  the  angle  BOG. 

.     Conclusions.    Angle  a  =  opp.  angle  a\ 

Angle  b  =  opp.  angle  b\ 
Proof.    1.  Because  the  sum  of  the  angles  BOD  and  DOA 
IS  the  angle  BOA,  and  OA  and  0^  are  in' a  straight  line. 

Angle  a  -f  angle  b  =  straight  angle.  (§§  18,'  61) 

2.  In  the  same  way  it  is  shown  that 

Angle  a'  -f  angle  b  =  straight  angle. 

3.  Comparing  (1)  and  (2), 

.Angle  a  -\-  angle  b  =  angle  a'  +  angle  b.  (§  44,  Ax.  1) 

4.  Take  away  from  these  equal  sums  the  common  angle  b 
and  we  have  ' 

Angle  a  =  angle  a'  {§  44,  Ax.  3).    Q.E.D. 
In  the  same  way  we  may  prove  that 

Angle  J  =  angle  J'.     Q.E.D. 

Theorem  II. 

68.  If  a  transversal  crossing  two  straight  lines 
makes  the  alternate  angles  equal,  the  two  straight 
lines  are  parallel. 

Hypothesis.  XY,  a  transversal  crossing  the  lines  A  B  end 
CD^t  the  points  M  and  N^  and  making  the  angle  ^  JfA^equal 
to  its  alternate  angle  MlSD. 

Conclusion.  AB\\  CD. 


29 


RBLATIOm  OF  ANGLES. 

Proof.  Bisect  the  length  MN  ski  the  point  P. 

Let  the  figure  be  turned 
half  way  round,  forming 
a  new  figure  with  the  ac- 
cented letters. 

Apply  the  new  figure  to 
the  old  one  so  that  the 
transversal  shall  be  turned 
end  for  end,  and  the  point 
JSf  shall  fall  upon  M, 
Then— 

1.  Because  JVM'=M]V, 
Point  M'  =  position  JV. 

2.  Because,  byhypoth., 
Ang.A'M'J!i'=  &iig.M^I), 
and 

Ang.  iriV2>'=ang.^  Jf JV, 

therefore  Line  M'A'~  trace  iVZ>; ) 

Line  JV'B'  =  trace  MaI\  (§  ^*) 

3  Therefore  the  whole  line  A'B'  will  fall  upon  CD,  and 
CD'  upon  AB  (§45,  Ax.  10). 

4.  Suppose,  if  possible,  that  the  lines  AB  and  CD,  when 
produced,  meet  in  the  direction  B  .and  D.  Then,  when  the 
figure  IS  inverted  the  liies  A'B'  and  C'D'  will  meet  in  the 
direction  B'  and  D\  Because  the  new  and  old  figures  coin- 
cide when  applied,^^  and  i>Cmust  also, when  produced,  meet 
m  the  direction  ^  and  Cas  well  as  in  the  direction  B  and  D 

5.  But  the  two  straight  lines  AB  and  CD  cannot  meet 
each  other  m  two  points  (§45,  Ax.  10,  Cor.). 

Therefore  they  do  not  meet  on  either  side. 

Therefore  they  are  parallel,  by  definition  (§  9).    Q.E  D 
^oroUary  1.  If  any  two  corresponding  angles,  as  C^Y  and 
AMJV,  are  equal,  then,  because  J/JVZ)  is  opposite  to  CNV  it 
IS  equal  to  It  (Th.  L),  and  the  alternate  angles  MND  and 
^^jYare  also  equal.     Hence— 

...^*^'  ^^^"^  i^'^^^^^^^rsal  crossing  two  straight  lines  makes 
any  two  corresponding  angles  equal,  those  lines  will  be  parallel 


1:1    il 


1  (  iu 
'I 

iil      ilii 


m  m 


I      i 


MSI  -!• 


30 


BOOK  IL    RECTILINEAL  FIGURES. 


70.  Corollary  2.  Any  two  perpendiculars  to  the  same 
straight  line  are  parallel. 

For  such  straight  hne  is  a  transversal  crossing  the  two 
perpendiculars,  and  making  the  angles  all  right  angles. 

Theorem  III. 

^1,  If  a  transversal  cross  two  parallel  straight 
lines,  the  four  alternate  and  corresponding  angles 
are  equal  to  each  other,  and  the  other  four  are  each 
equal  to  the  common  supplement  of  the  first  four. 


Hypothesis.  XF,  a  transversal  crossing  the  parallel  lines 
AB  and  CD  in  the  points  0  and  Q^  and  forming  with  them 
the  four  alternate  and  corresponding  angles 


a,    a',    a"f    a'", 


and  the  other  four  alternate  and  corresponding  angles 


h,    V,    b 


tt 


\nt 


.ttt 


\ttt 


Conclusions. 

1.  Angle  a  =  angle  a'  =  angle  «"  =  angle  a' 
II.  Angle  b  =  angle  b'  =  angle  A"  =  angle  b' 
III.  Any  angle  a  -f-  any  angle  b  —  straight  angle. 
Proof.  If  the  alternate  angles  a'  and  a"  are  not  equal, 
draw  through  0  the  line  A'B',  making  the  angle  ^'0^  equal 
to  its  alternate  angle  OQD.     Then — 

I.  Because  the  alternate  angles  are  equal. 

Line  yl'^' II  Hne  CD.    '  (§68) 

.    But  AB  II  CD,  by  hypothesis. 


(§67) 

Q.E.D. 
Q.E.D. 

(§51) 


RELATIONS  OF  ANGLES.  3^ 

Therefore  we  should  have  passing  through  0  two  straight 
hues  AB  and  A'B'  each  parallel  to  CD,  which  is  impossible 
(§  45,  Ax.  11).     Therefore 

Angle  a'  =  angle  a". 
Also,  Angle  a    =  angle  a'.  \ 

Angle  a'"  =  angle  a",  f 
Therefore 

Angle  a  =  angle  a'  =  angle  «"  =  angle  a'". 

II.  In  the  same  way  we  may  prove  that 
Angle  b  =  angle  b'  —  angle  b"  =  angle  *'". 

III.  Because  ^  0^  is  a  straight  line. 

Angle  a  +  angle  b  =  straight  angle.  ( 

But  all  of  the  four  angles  a  and  b  are  equal.     Therefore 
Any  angle  a  +  any  angle  b  =  straight  angle.    Q.E.D. 

72.  Corollary.     If  a  line  be  perpendicular  to  one  of  two 
parallels,  it  will  be  perpendicular  to  the  other  also. 

Theoeem  IV. 

73.  The  sum  of  the  three  interior  angles  of  a  tri- 
angle is  equal  to  a  straight  angle. 

Hypothesis.    ABC,  any  triangle. 

Conclusion.    Angle  A  +  angle  B  -f  angle  C  =  straight 
angle. 

Proof.  Through  C  draw  a  straight  line  MJV  parallel  to 
the  opposite  side  AB.     Then — 

1.  Because   CA    is  a  m _^ 

transversal   between    the  /s:         -  -     W 

parallels  AB  and  MN, 
Angle  J = alt.  angle  MCA 

(§  72). 

2.  Because   CB    is    a 
transversal  between  the  same  parallels, 

Angle  B  =  alternate  angle  BCJV. 

3.  Angle  C  =  angle  A  CB  (identically). 

4.  Adding  these  three  equations, 
Auffle^  4- anorlo  /?  4-  nno-lp  C—  Mf^  a  _]_  Arin   \    T>n\r 

~  angle  MCN, 
=  straight  angle. 


32 


BOOK  II,    RECTILINEAL  FIGURES. 


i  I 


Therefore 
Angle  A-\-  angle  B  +  angle  C  =  straight  angle.    Q,E.D. 

74,  Corollary  1.  If  two  angles  of  a  triangle  are  given, 
the  third  angle  mag  be  found  by  subtracting  their  sum  from 
180°. 

"75.  Corollary  2.  If  two  triangles  have  two  angles  of  the 
one  equal  respectively  to  two  angles  of  the  other,  the  third 
angles  will  also  be  equal. 

EXERCISES. 

1.  If  a  triangle  has  two  angles  each  equal  to  60°,  what 
will  be  the  third  angle? 

2.  If  one  angle  of  a  triangle  is  a  right  angle,  and  one  of 
the  remaining  angles  is  double  the  other,  what  will  be  the 
value  of  these  two  angles? 

3.  Prove  that  a  triangle  cannot  have  more  than  one  right 
angle. 

Theorem  V. 

76.  Each  exterior  angle  of  a  triangle  is  equal  to 
the  sum  of  the  two  interior  and  opposite  angles. 

Hypothesis.  ABC,  any  triangle.  /),  any  point  on  AB 
produced. 

Conclusion.     Exterior  angle  CBD  =  angle  A  +  angle  C. 

Proof.   1.  Because  ABB  are  in  one  straight  line, 

Angle  B  +  exterior  angle 
CBD  =  straight  angle. 

2.  Angle  B  +  angle  A  -f  an- 
gle C  =  straight  angle.      (§  73) 

3.  Cc':aparing  (1)  and  (2), 
Angle  A  +  angle  B  +  angle  C  =  angle  B  +  ext.  angle  CBD. 

4.  Taking  away  the  common  angle  B, 

Angle  A  +  angle  C  =  exterior  angle  CBD.     Q.E.D. 

77.  Corollary.  Any  exterior  angle  of  a  triangle  is  greater 
than  either  of  the  interior  and  opposite  angles. 

78.  T)ef.  Two  •nnrnllpl  linpa.  pa,r»h  DTiinoc  c\y\t  ^rc\xn  a 

point,  are  said  to  be  similarly  directed  or  oppositely 


lli 


RELATIONS  OF  ANGLES.  33 

directed  according  as  they  go  out  in  the  same  direc- 
tion or  in  opposite  directions  fioni  their  startine- 
points.  ° 

Theoeem  VI. 

79.  If  the  two  sides  0/ one  angle  are  respectwelv 
parallel  to  the  two  sides  of  another,  and  similarly 
directed,  these  angles  are  equal. 


A  S  «'^1  ^r?  """"f''  ^^^  ^'^^  ^^^  h^^i^g  the  sides 
a  J«n  f  ^ParaUel  and  similarly  directed,  and  the  sides ^P 
and^g  also  parallel  and  similarly  directed. 

Condmion.    Angle  JVBQ  =  angle  MAP. 

rroof.  Produce  the  side  JVB,  if  necessary,  in  either 
dn^ction  nntil  it  shall  intersect  the  side  AP  o  the  0  her 
angle,  also  produced  if  necessary. 

Produce  NB  past  B  to  any  point  8. 

Let  C  be  the  point  of  intersection  of  JSTB  and  AP.  Then— 

crosslnftrm!  ""^  "  '"'''''  ''  ""^^  '"'  ^  "^  "  ^  ''''''''''''' 
Angle  Jf^P  =  alternate  angle  ^C^.  (§  7i> 

Angle  iV5^  =  angle  A  OJS, 
3.  Comparing  (1)  and  (2), 

Angle  iV:^^  =  angle  Jf^P.    Q.E.D. 

the  other     BiiMh^^  ''  *^'  °°'  ^°^^"  ^«  ^^  ^^^  "<^t  <^o^tamed  within 
otner.    But  the  same  reasoning  can  be  applied  to  both. 

n.f^'  ^'?^^^^^^y  1-    Jf  t^e  sides  of  ttoo  anqhs  are  varalM 
and  oppositely  directed,  the  angles  will  l>e  equT        ^        '^ 


84 


BOOK  II.    RECTILINEAL  FIOUHES. 


.1 ,(, 


B 


81.  Corollary  2.  If  the  sides  are  parallel  and  the  one 
pair  are  similarly  directed  but  the  other  pair  oppositely 
directed,  the  angles  will  he  supplementary,  ^ 

Theorem  VII. 

S2.  TJie  Usectors  of  two  adjacent  angles  on  the 
same  straight  line  are  perp^ 
dicular  to  each  other.  \ 

Hypothesis.     BOO  smd  CO  A, 
adjacent  angles  on  the  straight  line 

^i?;  0 A,  OX,  the  bisectors  of  these  ^ 

angles.  ^ 

Conclusion.     OL  J.  0  lO,  that  is,  ZOK=  right  angle. 
Proof.     1.  By  hypothesis,  ^ 

Angle  COir=  |  angle  BOO. 

Angle  COL  =  ^  angle  COA. 
2.  Therefore 

Angle  A^OL  =  COR  +  COL, 

=  i{BOC-i-COA), 
=  i  straight  angle  BOA, 
=  right  angle.  Q  E  D 

Corollary.    Since  an  angle  can  have  but  one  bisector,  we 
conclude:  ' 

tn  ifh  ^/^'^V.^^^7^^,  ^^'  ^^^^^^  Of  an  angle,  perpendicular 
to  the  bisector,  bisects  the  adjacent  angle. 

Theorem  VIII. 

sa^fltrXMl^'  '^'"''  <^i'-.V.««^fe.  arc  intke 

Hypothesis.     AOL,  BOC,  two                           ^ 
opposite    angles    formed    by    the               >         / 
straight  lines  AB  and    CD;    OJ,                 ^ 
the  bisector  of  BOC;  OK,  the  Msec-     * 
tor  of  A  OB.  y^ B 

Conclusion.    OJ  and  OK  are  in        ^. 
the  same  straight  line. 

Proof.     1.  Because  the  angles 


RELATIONS  OF  ANGLES. 


35 


BO  J  and  AOK  mq  halves  of  the  equal  opposite  angles  BOGy 
AODy  they  are  equal  (§67). 

2.  Angle  AOJ  =  angle  AOB  -  angle  BO  J. 
Angle  KOJ  =  angle  AOJ  -\-  angle  yi Oif, 

=  straight  angle  AOB  -  BOJ-^AOK, 
Or,  30J  and  ^OA"  being  equal, 

Angle  KOJ  =  straight  angle, 
and  the  lines  O^and-  OJ  are  in  one  straight  line.     Q.E.D. 

86.  Corollary.  The  four  bisectors  of  the  four  angles 
formed  by  two  interseating  straight  lines  form  a  pair  of 
straight  lines  perpendicular  to  each  other. 

EXERCISES. 

1.  If,  in  the  diagram  of  Th.  II.,  angle  BMX  =  ^b°,  what 
will  be  the  values  of  the  other  seven  angles  of  the  figure? 

2.  If,  in  the  diagram  of  Th.  VIII.,  angle  BOJ  =  20°,  com- 
pute the  values  of  the  angles  CO  Ay  AOD,  and  DOB. 

3.  Draw  a  triangle  ABCy  and  suppose  angle  A  =  60°,  angle 
B  =  40°.  What  angle  will  the  bisectors  of  these  angles  A  and 
B  form  with  each  other? 

Note.  Apply  Th.  IV.  to  the  triangle  formed  by  the  bisectors  and 
the  side  AB. 

4.  What  will  be  the  values  of  the  three  exterior  angles  of 
the  preceding  triangle  4BC? 

5.  If,  in  the  preceuing  triangle  ABO,  the  bisector  of  the 
angle  A  be  continued  until  it  cuts  the  side  BC,  what  angles 
will  it  form  with  that  side? 

6.  Compute  the  same  angles  algebraically  in  terms  of  the 
angles  AyBy  and  (7,  and  show  that  the  difference  between  the 
two  adjacent  angles  formed  by  the  bisector  with  the  side  BO 
is  equal  to  the  difference  between  the  angles  B  and  C. 

7.  If,  in  a  triangle  ABO,  exterior  angle  A  =  85°  and  ex- 
terior angle  B  =  150°,  what  will  be  the  three  interior  angles? 


if 


II 


IP 


■  i; 


mi 


llr 


I  i  III 


CHAPTER    II 

REUTIONS  OF  TRIANGLES. 


Equilateral 
triangle. 


Isosceles 
triangle. 


Definitions. 

86.  Definition.    An  equilateral  triangle  is  one  in 
which  the  three  sides  are  equal. 

Def.  An  isosceles  triangle 
is  one  which  has  two  equal 
sides. 

Def.  An  acute-angled  tri- 
angle is  one  which  has  three 
acute  angles. 

Def.    A  right-angled  triangle  is  one  which  has  a 
right  angle. 

Def.    An   obtuse-angled    triangle  is  one  which 
has  an  obtuse  angle. 

87.  Def  In  a  right- 
angled  triangle  the  side 
opposite  the  right  angle 
is  called  the  hypotlie- 
nuse. 


Blght-angled 
triangle. 


Obtuse-angled 
triangle. 


88.  Def  When  one 
side  of  a  triangle  has  to  be  distinguished  from  the 
other  two  it  is  called  a  base,  and  the  angle  opposite 
the  lase  is  called  the  vertex. 

Either  side  of  a  triangle  may  be  taken  as  the  base,  but  we 
commonly  take  as  the  base  a  side  which  has  some  distinctiTe 
property. 

In  an  isosceles  triangle  the  base  is  generally  the  side  which 
is  not  equal  to  another. 

In  other  triangles  the  base  is  the  side  on  which  it  is  sup- 
posed to  rest. 


IS  one  m 


RELATIONS  OF  TRIANGLES.  37 

89.  i>(?/.    An  obUque  line  is  one  which  is  neither 
perpendicular  nor  paraUel  to  some  other  Hne. 

90.  JD^    Segments  of   a  straight  line  are  the 
parts  into  which  it  is  divided. 

Theorem  IX. 

91.  In  an  isosceles  triangle  the  angles  opposite  the 
equal  sides  are  equal  to  each  other. 

Hypothesis.   ABC  a,  triangle  in  which 

CA  =  CB. 
Conclusion.  Angle  A  =  angle  B. 
Proof.  Bisect  the  angle  C  by  the  line 
OD,  meeting  AB  in  B.     Turn  the  tri- 
angle oyer  on  CD  as  an  axis.     Then— 
1.  Because,  by  construction, 
Angle  BCD  =  A  CD, 

Side  CA  =  trace  CB,  and  vice  versa. 
3.  Because,  by  hypothesis,  CA  =  CB, 
Point  B  =  position  A, 
^^^  Point  ^E  position^. 

3.  Therefore  line  AB  =  traxie  BA,  being  turned  end  for 
end. 

4.  Therefore  angle  CU^e  trace  CBA. 

5.  Therefore  angle  CAB  =  angle  CBA  (§  14).     Q.E.D. 
Corollary  1.    Since,  after  being  turned  over,  the  triangle 

lalls  upon  its  own  trace,  the  triangle  is  symmetrical  with 
respect  to  the  bisecting  line.     Hence— 

93.  The  bisector  of  the  vertical  angle  of  an  isosceles  tri- 
angle Is  an  axis  of  symmetry,  and  bisects  the  base  at  right 
angles.  ^ 

93.  Cor.  2.  Every  equilateral  triangle  is  also  equiangular. 

Theorem  X. 

94.  Conversely,  if  two  angles  of  a  trianale  are 
mal,^  the  sides  opposite  these  angles  are  equaL  and 

trie  tnanale  i.t  iananplfis^ 


the  triangle  is  isosceles. 


38 


BOOK  11.    REOTILINEAL  FIQUBE8. 


Hypothesis.  ABC,  a  triangle  in  which  angle  A  =  angle -ff. 

Conclusion.   Side  CA  =  side  CB. 

Proof.  Through  the  middle  point  /> 
of  the  side  AB  pass  a  perpendicular,  and 
turn  the  figure  over  upon  this  perpen- 
dicular as  an  axis.     Then — 

1.  Because  the  axis  bisects  AB  per- 
pendicularly, 

Point  A  =  position  B.  )        ,„  .q. 
Point  B  =  position  A.  )        ^»  ^^ 
Therefore  AB  =  trace  BA. 

2.  Because  angle  A  =  angle  B, 

Side  AC- trace  BC 
Side  BC=tTace  AC. 

3.  Therefore  the  point  of  intersection  C7will  fall  into  its 
original  position. 

4.  Therefore  ^C=jrC.     Q.E.D. 

95.  Corollary  1.  A  line  Usecting  the  hose  of  an  isosceles 
triangle  perpendicularly 

Passes  through  its  vertex, 
Is  an  axis  of  symmetry,  and 
Bisects  the  angle  opposite  the  base. 

96.  Cor.  2.  Bvery  equiangular  triangle  is  also  equilateral 

Theorem  XI. 

97.  If  in  any  triangle  one  side  be  greater  than 
another,  the  angle  opposite  that  side  will  he  greater 
than  the  angle  opposite  the  other. 

Hypothesis.     ABC,  Vk  triangle 
in  which 

AB>  BC>  CA. 

Conclusion. 
Angle  C  >  angle  A  >  angle  B. 

Proof.  On  a  greater  side,  CB, 
take  CD  equal  to  a  lesser  side,  ^^ 
CA,  and  join  AD.     Then — 

1.  Because  the  line  AD  falls  within  the  triangle, 
Angle  CAB  >  angle  CAD. 


HELATIONS  OF  TRIAKGLES, 
2.  Because  CA  =  CDy 

Angle  CAD  =  angle  CDA. 


89 


(§03) 


3.  Because  CDA  is  an  exterior  angle  of  'the  triangle  AlS 
A   1?        ^^^^^/^ODA>  angle  ABD,  (iii\ 

from  Ir  ^^         ^'^  "'  '"'  "^'^  '^^^  >  ^^^'  -S 

Angle  GAB  >  angle  ^^Z>. 
m  tne  same  way  may  be  shown. 

Angle  C>  angle  ^.     Q.E.D. 

Theorem  XII. 
98.   CouTiersely,  if  one  angU  o^  a  triavaJp  h. 

Hypothesis.    ABC,  a  triangle  in  which 

-^^gle  O  angle  ^  >  angle  ^.       ' 

^^Proof.  Prom(7draw  CD,  making  angle  ACD  =  angle  (7^  A 

1.  Because  angle  ACD  z=  angle  CAD, 

AD  =  CZ>.  '  /£>  q;.x 

or,  from  (1), 

AB^CD  +  DB, 

3.  Because  CB  is  a  straight  J 
hne,  ^     ^ 1^ -^B 


4.  Therefore,  from  (2), 

0.  in  the  same  way  may  be  shown 

BC>AC.     Q.E.D. 


(Ax.  9) 


the  order  ofmagmtude  of  their  opposite  angles. 


the 


40 


BOOK  11.    liEOTILINEjiL  FIOUIUCS. 


That  Is,  if 

Angle  A  >  angle  B  >  angle  C,  then 
Side  BC>  side AG>  side  AB, 
and  vice  versa. 

Theorem  XIII. 

100.  If  from  any  point  within  a  triangle  lines  he 
drawn  to  the  ends  of  the  hase^  the  sum  of  these  lines 
will  he  less  titan  the  sum  of  the  other  two  sides  of  the 
triangle,  hut  they  will  contain  a  greater  angle. 

Hypothesis.     ABC,  any  triangle;  P,  any  point  within  it. 
Conclusions.     I.  AF -{■  FB  <  AC -^  CB. 

II.  Angle  AFB>  angle  ACB. 
Froof.     Continue  the  line  AF  until  it  meets  the  side  CB 
in  Q.  Then— 
(1)1.  AQ<AC-{-CQ.{Ax.9) 
2.  Adding  QB  to  both  sides 
of  this  inequality,  we  have 
AQ -{■  QB  <  AC -\- CQ -}- QB; 

(Ax.  6) 
that  is,  A." 

AQ-\-QiJ<AC-^CB. 

3.  Abo,  in  the  same  way, 

FB<FQ-^QB.  (Ax.  9) 

4.  Adding  ^  P  to  both  sides, 

AF  +  FB<AF  +  FQ-\-  QB, 
or  AF  +  FB  <  AQ -\- QB. 

Comi)aring  (4)  and  (2), 

AF  +  FB  <AC+CB.     Q.E.D. 
(II)  5.  Because  FQB  is  an  exterior  angle  of  the  triangle 
ACQ, 

Angle  FQB  >  angle  A  CB. 

6.  Because  AFB  is  an  exterior  angle  of  the  triangle  FQB, 

Angle  AFB  >  angle  FQB, 

7.  Comparing  (5)  and  (6), 

Angle  AFB  >  angle  A  CB.    Q.E.D. 


:iih::ii 


RELATWm  OF  TRIAmLES. 


41 


Theorem  XIV. 

101.  HVom  a  point  outside  a  straight  line  only 
one  perpendicular  can  be  drawn  to  such  straiqht 
Line,  and  this  perpendicular  is  the  shortest  distance 
from  the  point  to  the  line. 

Hypothesis.  AB,  any  line;  P,  any 
point  without  it;  PO,  u  perpendicu- 
liir  from  P  on  AB]  PQ,  my  other  line 
from  P  to  A  B. 

Conclusions. 

I.  PQ  is  not  perpendicular  to  AB. 
II.  PO  <  PQ. 

Proof.  Turn  tlie  figure  over  on 
AB  m  [in  axis,  so  that  the  jmint  P 
shall  fall  into  the  position  P'.    Then— 

1.  Because  AOP  is  a  right  angle, 
AOP'  is  also  a  right  angle,  and 
POP'  lie  in  a  straight  line  (§  60,  Cor.). 

2.  If  PQA  were  also  a  right  angle,  it  could  be  shown  in 
the  same  way  that  PijP'  is  a  straight  line,  and  there  would 
be  two  straight  Imes  POP'  and  PQP'  having  the  points  P 
and  P  common,  which  is  impossible  (§  45,  Ax.  10,  Cor.). 
Iherefore  PQA  is  not  a  right  angle.     Q.E.D 

3.  Because  PO  =  OP',  and  PQ  =  QP%  we  have 

PO  =  iPP';PQ  =  ^^PQ.^Qp.^^ 

But  pp'  <PQ-{-  QP'.  (8  45  Ax  9^ 

Therefore,  taking  the  half  of  these  unenual  quantities,' we 
have  PO  <  PQ.     Q.E.D. ' 

Theoeem  XV. 
103.  Two  oblique  lines  from  a  point,  cutting  a 
straight  line  at  equal  distances  from  the  f      of  the 
perpendicular,  are  equal  in  length,  and  man.,  \qual 
angles  with  the  line.  ^ 

Hypothesis.    AB,  any  straight  line;  P,  any  point  outside 
It;  PO,  the  perpendicular  from  P  to  AB,  meetincr  ABinG 
M,  iV;  two  points  on  the  line  AB,  such  that  0M^=  0N\    ^ 


V\ 


I 


f 


, 


I  1 


42 


BOOK  II    RECTILINEAL  FIQUBES. 


Conclusions, 


I.  PN=PM. 
II.  Angle  PMO  =  angle  PNO, 
Proof.    Turn  the  figure  over  on  PO  as  an  axis.    Then— 

1.  Because    angle  POM  =  p 
angle  POiV^  (hypothesis), 

O^E  trace  OB. 

2.  Because  OM  =z  ON, 

Point  M=  JSr. 

3.  Therefore     PM  =  PJV, 
and  angle  PMO  =  angle  PiVO. 

Q.E.D. 


Theorem  XYI. 

103.  (y  two  oblique  lines  drawn  from  a  point  to 
a  straight  line,  that  which  meets  the  line  at  the  greater 
distance  from  the  foot  of  the  perpendicular  is  the 
longer,  and  makes  the  lesser  angle  with  the  line. 

Hypothesis.  AB,  any 
straight  line  ;  P,  any  point 
without  it ;  PO  perpendicular 
to  AB  \  Q,  R,  any  two  points 
on  AB,  such  that  0R>  OQ. 

Conclusions.  I.  PR  >  PQ. 
II.  Angle  PRO  <  angle  PQO. 

Proof.  Turn  the  figure  over 
on  ^P  as  an  axis.  Let  P'  be 
the  point  on  which  P  shall 
fall.     Then— 

1.  Because  P'Q  =  PQ,  and  P'P  =  PR, 

PQ  =  UPQ  +  QP'). 
PR  =  \{PR-\-RP'). 

2.  If  Q  is  on  the  same  side  of  0  with  R,  the  point  Q  will 
be  within  the  triangle  PRP'.     Therefore 

ppj-ppf'-^pnj-np' 

and,  taking  the  halves  of  these  unequal  quantities. 


PR  >  PQ.     Q.E.D. 


RELATIONS  OF  TRIANQLB8. 


48 


3.  If  ^  is  on  the  opposite  side  of  0  from  R,  take  another 
point  Q  on  the  same  side. 

The  two  lines  PQ  wiP  then  be  equal  to  each  other  (§  102). 
The  second  point  Q  will  fall  between  0  and  i2  (hypoth.). 

4.  Therefore  we  shall  still  have 

PR  >  PQ.     Q.E.D. 

5.  Because  PQO  is  an  exterior  angle  of  the  triangle  PQR, 

Angle  PRQ  <  angle  PQO  (§  77).    Q.E.D. 


I 


9 


Theoeem  XVIL 

104.  I.  Mery  point  on  the  perpendicular  bisector 
of  a  straight  line  is  equally  distant  from  the  extremi- 
ties of  the  line. 

II.  Every  point  not  on  the  perpendicular  bisector 
is  nearer  that  extremity  t(mard  which  it  lies. 

Hypothesis.  AB,  a  straight 
line;  0,  its  middle  point;  OP,  a 
perpendicular  from  0;  P,  any 
point  on  this  perpendicular;  Q, 
a  point  on  the  same  side  of  the 
perpendicular  with  B. 

Conclusions.    I.  PA  =  PB. 

II.  QB  <  QA.       / 

Proof.  1.  Because  PO 1.  AB,  A^ 
and  ^  0  =  OBf  we  have 

PA  =  PB.  (§  102) 

2.  From  Q  drop  a  perpendicular  QO'  upon  AB.     Then 

QO'  II  PO.  (§  70) 

Therefore  0'  falls  on  the  side  of  0  toward  B,  and 

O'B  <  0A\ 
Therefore  QB  <  QA  (§  103).    Q.E.D. 

105,  Corollary.  Every  point  equally  distant  from  the 
extren.  :Hes  of  a  line  lies  upon  the  perpendicular  bisector  of 
the  line. 

For,  if  it  did  not  lie  on  this  hisenf.nr  if.  nnniA  r./xf  r^« 
equally  distant  from  the  extremities  without  violating  conclu- 
sion II.  of  the  theorem. 


1 


•.. 


O' 


\  \ 

:i 


B 


I' I 
m 


1 


11,     17 


m 

liliii/i 


44  BOOK  II.    RECTILINEAL  FIGURES. 


Theorem  XVIII. 

106.  EGery  point  in  the  bisector  of  an  angle  is 
equally  distant  from  the  sides  of  the  angle;  and  every 
point  within  the  angle,  hut  not  on  the  Usector  is 
nearer  that  side  toward  which  it  lies.  ' 

Hypotlitsis,  MON,  any  angle;  and  OQ,  its  bisector,  so 
that  angle  MOQ  =  NOQ ;  P,  any 
point  on  OQ-,  T,  a  point  within  the 
angle,  but  not  on  OQ-,  PR,  FS,  per- 
pendiculars to  OM  and  ON;  TU, 
TV,  perpendiculars  from  T  to  OM 
and  OJV^. 

Conclusions.    I.  PR  =  PS. 
11.  TU>  TV. 
Proof  I.  Turn  the  figure  over  on  the  axis  OP.    Then— 

1.  Because  angle  PON=  POM, 

Line  0M=  trace  ON,  and  vice  versa. 

2.  Because  PR  is  a  perpendicular  dropped  from  P  on  R, 

PR  =  trace  PS.  (Th  101) 

3.  Therefore  Point  R  =  S,md.  \     •       J 

PR  =  PS.    Q.E.D. 

Proof  II.  Let  Q  be  the  point  in  which  TU  crosses  the 
bisector  OQ.  From  Q  drop  the  perpendicular  QW  nvon  ON 
Join  TW.    Then— 

4.  Because  TV  is  a  perpendicular  and  TWm  oblique  line 
to  ON, 

TW>TV.  (§101) 

5.  Also,  TQ^QW>TW.  (Ax.  9) 
Or,  because  QU=QW  (I.),  ^         ^ 

TQ^QU=TU>TW. 

6.  Comparing  with  (4), 

TU>  TV.    Q.E.D. 

107.  Corollary.  Every  point  equally  distant  from  two 
non-parallel  lines  in  the  same  plane  lies  on  the  Usector  of  the 
angle  formed  hy  those  lines. 


RELATIONS  OF  TBIAmLEa. 


45 


Theorem  XIX. 

.7.  i^^*  ^  two  triangles  have  two  sides  and  the  in- 
eluded  angle  of  the  one  equal 
to  two  sides  and  the  included 
angle  of  the  other,  they  are 
identically  equal. 

Hypothesis,  -4.5 (7 and  MNP, 
two  triangles  in  which 
PM^  CA. 
PJ^=  CB. 
Angle  P  =  angle  O. 
Conclusion.     The    two    tri- 
angles are  identically  equal. 

H  SZ'-^'    .^^^^J  *.^'  ^""^^^^^^  ^^^  *«  ^^O  in  such  manner 
tha  the  vertex  P  shall  fall  on  C,  and  PM  on  CA.  Thr~ 
1.  Because  PM  =  CA, 

Point  if  =  points. 
3.  Because  angle  P  =  angle  C, 

Side  PN~CB.  /g  1,^ 

3.  Because  PN=  CB,  ^^     ' 

Point  JV^=  point  5. 

4.  Because  M=A  and  N=  B, 

Line  MN~  line  AB.  (§  45,  Ax.  10,  Cor. ) 
Therefore  every  part  of  the  one  triangle  will  coincide 
with  the  corresponding  part  of  the  other,  and  the  two  trT 
angles  are  identically  equal  by  definition  (§13).     Q.  E.  D. 

Theorem:  XX. 

.^/^^;  ^  ^T  ^l^'^^'Oles  haw  a  side  and  the  two 
adjacent  angles  of  the  one  equal  to  a  side  and  the 

tual  '       ^^'  ''^^'^'  ^^''^  ""^^  ^^e7^^^ca% 

Hypothesis.  ABCmd.  MNP,  two  triangles  in  which 

AB  =  MJV. 


a  »%  /v  I  r\       A     —  —  — 1  _       H  .T" 


Conclusion. 


Angle  B  =  angle  N. 
The  two  triangles  are  identically  equal. 


46 


BOOK  II.    RECTILINEAL  FIQUBES. 


Proof.     Apply  the  triangle  ABO  to  the  triangle  MNP  in 
such  manner  that  A  shall  coin- 
cide with  Mf  and  AB  with  MN, 
Then— 

1.  Because  AB  =  MNj 

Point  B  =  point  iV^. 

2.  Because  angle  ^  =  angle  M, 

Side  AC  =  aide  MP.  ^^ 

3.  Because  angle  B  =  angle  JV", 

Side  J?(7=  side  iVP. 

4.  Because  the  sides  A  0  and 

^C  fall  upon  MP  and  iVP  respectively,  the  vertex  C  will  fall 
upon  the  vertex  P. 

Therefore  the  two  triangles  coincide  in  all  their  parts  and 
are  identically  equal.  .  Q.E.D. 


Theorem  XXI. 

110.  If  two  triangles  have  the  three  sides  of  the 
one  respectively  equal  to  the  three  sides  of  the  other, 
they  are  identically  equal,  and  have  the  angles  op- 
posite the  equal  sides  equal. 

Hypothesis.  Two  tri- 
angles, ABCm.dDEF, 
in  which 

AB  =  DK 
BG  =  EF, 
CA  =  FD. 

Conclusion.  The  two 
triangles  are  identically 
equal. 

Proof.  Take  up  the  triangle  ABC  and 
apply  the  side  ^i5  to  the  equal  side  DE  oi 
the  other  triangle,  letting  the  vertex  C  fall 
on  the  opposite  side  of  DE  from  that  on 
which  the  triangle  DEF  lies. 

Let  C"  be  the  point  in  which  the  vertex  C  falls. 

Join  FC.     Then— 

1.  Because  FD  ==  AC,  wid  DC  —  AC,  it  follows  that 


!/ 


%' 


RELATIONS  OF  TRIANGLES. 


47 


FD  =  DC,  and  the  triangle  FBC  is  isosceles.    Therefore 
Angle  DFC  =  angle  DC'F,  (§  91) 

2.  For  the  same  reason  the  triangle  FFG'  is  isosceles,  and 

Angle  BFC  =  angle  FC'F. 

3.  Adding  the  equations  (2)  and  (1),  we  find 

Angle  BFC  +  angle  HFC  =  angle  I)C'F-\-  angle  BC'F. 
But  Angle  DFO'  +  angle  BFO'  =  angle  i9i^^. 

Angle  DC'F-{-  angle  EC'F=  angle  />C"^. 
Therefore 

Angle  DC'iS'  =  angle  DFE. 

4.  Butangle^CJ5  =  angle  Z)C"^,  by  construction.  There- 
fore 

Angle  ACB  =  angle  i)i?!£'. 

5.  The  two  giyen  triangles,  having  the  angle  C=  angle  F 
and  the  sides  which  contain  these  angles  equal,  are  identically 
equal  (§108).     Q.E.D.  ^ 

Theoeem  XXII. 

111.  If  two  triangles  agree  in  the  lengths  of  two 
sides  and  also  in  the  angle  opposite  one  of  these  sides, 
the  angles  opposite  the  other  of  the  equal  sides  will  be 
either  equal  or  supplementary,  and  if  they  are  equal 
the  triangles  are  identically  equal. 


BD 


E  D 


E        P 

Hypothesis,    Two  triangles  ABC  and  DBF,  in  which 

CA  =  FD. 
CB  =  FK 
Angle  A  =  angle  D, 
Conclusion.   Either  angle  E  =  angle  B  or 

Angle  E  =  straight  angle  —  angle  B, 

and  in  thfl  formfir  ostap.  t.liA  twn  frianorlfifl  aro  irlAyifi/Jollir  AiNnol 

Proof.    Apply  the  triangle  DEF  to  the  triangle  ^^C  in 
such  manner  that  DF  shall  fall  on  the  equal  side  A  C.  Then 


i|; 


'h 


48 


BOOK  II.     nECTILINEAL  FIGURES. 


Kl 


\ 


B  D 


E  D 


1.  Because  i>i^=  ^C, 

PointDE^;  point  i^E  a 

2.  Because  angle  D  =  angle  A, 

Base  I) E= base  A  B. 

3.  Because  F.E  =  CB,  the  point  B  will  fall  on  a  point  of 
the  base  AB  which  is  at  a  distance  from  C  equal  to  CB, 

4.  There  will  be  two  such  points  equally  distant  from  the 
foot  P  of  the  perpendicular  from  C  on  AB  (§  103). 

Because  FF!  =  CB,  one  of  these  points  will  be  B.    Let  B' 
be  the  other. 

5.  If  iSf  falls  on  5, 

Triangle  ABC=  triangle  DBF,  identically. 

6.  If  B  falls  on  B',  then,  because  CB'  =  CB,  the  triangle 
CB'B  will  be  isosceles,  and 

Angle  CB'P  =  angle  CBP,  (§91) 

7.  Because  AB'P  is  a  straight  line. 

Angle  CB'A  =  supplement  of  angle  CB'B, 

=  supplement  of  angle  CBP.  (6) 

8.  But  in  this  case  angle  CB'A  =  angle  B.     Therefore 

Angle  B  —  supplement  of  angle  CB'P, 

=  supplement  of  angle  B  (7).    Q.E.D. 

113.  Corollary.    If  the  triangle  ^5C  should  be  right- 
angled  at  B,  we  shall  have 

Angle  B  =  straight  angle  —  angle  B, 
and  the  two  possible  angles  B  would  have  the  same  value. 
Hence  the  two  triangles  would  then  be  identically  equal. 

Scholium. 

113.  The  nrecedinsr  theorems  of  the  idftutitv  of  triaticyl«g 
are  also  expressed  by  saying  that  when  certain  parts  of  a 


RELATIONS  OF  TRIANQLES. 


49 


A  hinged  triangle. 


triangle  are  given,  the  other  parts  are  determined.  T>he 
parts  of  a  piano  triangle  are  the  three  sides  and  the  three 
angles. 

The  three  angles  ai'e  not  all  independent,  because  whenever 
two  of  them  arc  given  the  third  ftiay  be  found  by  subtracting 
their  sum  from  a  straight  angle  (§  74). 

Whenever  three  independent  parts  of  a  triangle  are  given 
the  remaining  parts  may  be  found.    In  other  words,  when 
three  independent  parts  are  given  there  is  only  one  triangle 
(or,  in  the  case  of  §  111,  two  triangles)  having  those  parts. 

When  the  three  sides  of  a  triangle  are  given,  we  may  imagine 
ourselves  to  have  three  stiff  thin  rods  which  we  can  fasten 

end  to  end  in  the  form  of  a  tri- 
angle. When  the  angles  are  not 
given,  we  may  suppose  the  rods  to 
be  fastened  together  by  hinges  at 
the  angles. 

Theorem  XXI.  shows  that  al- 
though  the  hinges  may  be  quite 
free  the  rods  cannot  turn  upon  them  when  linked  together. 
If  they  could  turn,  we  could  make  the  rods  into  several 
triangles  by  turning  the  rods  on  the  hinges,  and  these  tri- 
angles would  not  be  identically  equal. 

When  two  sides  and  the  included  angle,  as  AC,  BO,  and 
the  angle  (7  are  given,  which  is  the  case  corresponding  to 
rheoi;em  XIX  we  must  suppose  the  side  AB  removed  and 
the  hinge  at  C  tightened,  so  that  the  two  rods  cannot  turn, 
and  we  are  required  to  find  a  third  rod  of  such  length  as 
to  fi  into  the  space  AB.  Theorem  XIX.  shows  that  this  rod 
must  have  a  definite  length. 

Suppose  next,  as  in  Theorem  XXIL,  that  AC,  BC,  md 
the  a^gle  A  are  given,  while  the  base  AB  and  the  angles  B 
and  C  are  not  given.  We 
may  then  suppose  a  long 
rod  extending  out  from  A. 
The  side  AC  of  given 
length  must  be  fastened  at 
A,  and  the  hinge   tight-  ^  „ 

ened  so  that  AG  cannot  turn,  becaule  the  angle '^  is  given. 


Bf'--^  -- 


f'     ! 


50 


BOOK  II.    RECTILINEAL  FIGURES. 


,i  ^i 


Wi  m 


X 


/ 


The  rod  CB  of  given  length  is  hinged  at  C,  and  this  hinge  is 
left  loose  because  the  angle  C  is  not  given.  We  are  then  to 
swing  the  side  CB  around  on  0  until  the  end  B  touches  the 
base,  when  it  is  to  be  fastened  and  the  angle  well  fixed. 
There  will  be  two  points,  B'  and  B',  where  the  junction  may 
be  made,  and  only  two.  We  may  choose  which  point  we  will, 
and  the  triangle  will  then  be  fixed.  The  two  angles  at  B  will, 
by  the  last  theorem,  be  supplemen- 
tary. 

If   CB  should  be  shorter  than 
the  perpendicular    from     C  upon  /-        \ 

AB,  there  would  be  no  triangle 
which  could  be  formed  from  the 
given  parts.  A 

SuppoF;e,  lastly,  thaj;  one  side  AB  and  the  two  adjacent 
angles  A  and  B  are  given,  the  other  two  sides  being  of  in- 
definite length.  We  must  then  turn  the  two  sides  on  the 
hinges  and  tighten  the  latter  at  the  required  angles,  when  the 
sides  will  cross  each  other  at  a  definite  point,  and  will  make 
a  definite  angle  with  each  other.  This  corresponds  to  the 
case  of  Theorem  XX. 


^ 


B 


i'l 


Theoeem  XXIII. 

114.  If  two  triangles  agree  in  the  length  of  two 
sides,  that  triangle  in  which  these  two  sides  include 
the  greater  angle  will  have  the  greater  base. 

Hypothesis.     ABC&nd.  DBF,  two  triangles  in  which 

AB  =  DE, 
BC  =  BF, 
Angle  B  <  angle  B. 
Conclusion.     Ba.8e  DF>  hase  AC. 
Proof.    Apply  the  side  ^^  of  the  one  triangle  to  the 
equal  side  DF  of  the  other  in  such  manner  that  B  shall  fall 
upon  F,  and  A  upon  D.    Let  C  be  the  position  in  which  G 
falls. 

Bisect  the  angle  C'FF,  and  let  iVbe  the  point  in  which 
the  bisector  meets  the  base  DF.    Join  C'iV.     Then— 


RELATIONS  OP  TRIANGLES. 

1.  In  the  two  triangles  C'^JVand  FEN, 

Angle  NEC  =  angle  NEF,  by  constrnction. 
Side  EC  =  side  EF,  by  hypothesis. 
Side  E2i  =  side  EN,  identicaUy. 


01 


Therefore  triangle  ENC  =  triangle  ENF,  identically,  (S 110) 
and  NG'  =  NF. 

2.  Therefore  DF=  DN-\-  NF=  DN-\-  NG\ 

3.  Because  ^  67  is  a  straight  line, 

I)N+NO'>DO\  (Ax  9) 

Comparing  (2)  and  (3),  *    ' 

DF>Da% 
®r  DF>AO.     Q.E.D. 

115.  Corollary.  Conversely,  if  two  triangles  have  two 
sid^softhe  one  equal  respectively  to  two  sides  of  the  other,  hut 
the  third  sides  unequal,  the  angle  opposite  the  greater  of  the 
unequal  sides  will  he  the  greater. 

For  these  angles  could  not  be  equal  without  violating 
IheoremXIX.,  nor  could  the  angle  opposite  the  lesser  side 
De  greater  without  violating  Theorem  XXIII. 


Theorem  XXIV. 

IX^.  If  three  or  more  lines,  maJcing  equal  angles 
mm  each  other,  he  drawn  from  a  point  to  a  straight 
tine,  tji'^tmir^  of  lines  will  intercept  the  greater  length 
which  IS  farther  f Torn  the  perpendicular. 

Hypothesis.  PC,  a  straight  line;  0,  any  point  outside  of 


52 


BOOK  II.    RECTILINEAL  FIOUBES. 


\ 


it;  OAy  OB,  OGy  three  lines  from  0  to  PC,  making  angle 
A  OB  =  angle  BOC\  OP,  the  perpen- 
dicular from  0  upon  P, 

Conclusion.  The  intercept  BG,  on 
the  side  of  OB  away  from  the  perpen- 
dicular, will  be  greater  than  the  inter- 
cept AB,  on  the  side  toward  the  per- 
pendicular. 

Proof,  From  B  draw  the  line  BSy 
making  angle  OBS  —  angle  OBA,  and 
meeting  OG'in  S.    Then — 

1.  Because,  in  the  triangles  OBA  and  OBS, 

Angle  BOA  =  angle  BOS  (hypothesis), 
Angle  OBA  =  ajigle  OBS  (construction), 
Side  OB  =  side  OB  identically, 
these  triangles  are  identically  equal  (§  109),  and 

Angle  OSB  (opp.  OB)  =  angle  OAB  (opp.  OB). 
Side  BA  =  side  BS. 

2.  Therefore  angle  GSB  (supplement  of  OSB)  =  angle 
OAP  (supplement  of  OAB). 

3.  Because  G  is  farther  from  P  than  A  is. 

Angle  SGB  <  angle  OAP. 
Therefore       Angle  SGB  <  angle  GSB, 
and  side  BG  (opposite  greater  angle  GSB)  >  side  BS  (oppo- 
site lesser  angle  SGB).  (§  98) 

4.  Comparing  with  (1), 

BG>AB.    Q.E.D. 


(§  103) 


PARALLELS  AND  PARALLKLOOIiAMS. 


63 


CHAPTER    III. 

PARALLELS  AND  PARALLELOGRAMS. 


Definitions. 

117.  Def.    A  quadrilateral  is  a  figure  formed 
by  four  straight  lines  joined  end  to  end. 

The  sides  of  a  quadrilateral  are  the  lines  which 
form  iii. 

118.  Def.  A  parallelogram  is  a  quadrilateral  in 
which  the  opposite  sides 
are  parallel. 

Whenever  two  parallels 
cross  two  other  parallels,  the 
intercepted  portions  of  the 
parallels  form  a  parallelogram. 

119.    Def.     ThediagO-  a  parallelogram. 

nals  of  a  quadrilateral  are  two  lines  joining  its  oppo- 
site  angles. 

Theorem  XXV. 

120.  Straight  lines  which  are  parallel  to  the  same 
straight  line  are  parallel  to  each  other. 

Hypothesis.  The  line  i  par- 
allel to  the  line  a.  The  line  c 
also  parallel  to  the  line  a.  a  — 

Conclusion.  The  lines  i  and 
c  are  parallel  to  each  other.  5 

Proof.  Draw  any  transver-  / 

sal  as  ifiV^  across  the  three  lines,  " " " —70 

intersecting  them  in  the  points  / 

A,  B,  a  N 

1.  Because  h  is  parallel  to  a, 

Angle  5  =  corresponding  angle  ^.  (871) 

Angle  O  =  corresponding  angle  A, 


M 

/ 


/ 


/ 

IT 


54  I^OOK  IT.     HEGTIUNEAL  FlQUUm. 

3.  Comparing  (1)  and  (2), 

Anglo  B  =  corresponding  angle  C 

4.  Therefore  line  b  ||  lino  c  (§  69).     Q.E.D. 


Theorem  XXVI. 

121.  The  opposite  angles  of  a  parallelogram  are 
equal  to  each  other. 

Hypothesis.  A  BCD,  any  parallelogram. 
Conclusion, 

Anglo  A  =  opposite  angle  D. 
Angle  B  =  opposite  angle  G. 
Proof.  Continue  CD  to  any  point  Jf,  and  BD  to  any 
point  N.  js 

Then—  \ 

1.  Because  DN  is  parallel  to 
AC  and  similarly  directed,  and 
DM  parallel  toAB  and  similarly 
directed, 

Angle  BAC=  angle  MDN.  (§ 79) 

2.  Angle  MDN  =  opp.  angle  BBC,  (§  67) 

3.  Comparing  (1)  and  (2), 

Angle  BDC=  angle  BA  O. 
In  the  same  way  it  may  bo  proved  that 

Angle  ^CZ>  =  angle  ^^/).    Q.E.D. 


-M 


Theorem  XXVII. 

123.  Anp  two  adjoining  angles  of  a  parallelo- 
gram are  supplementary. 

Proof.  AiLj  such  pair  of  angles  as  A  and  B  are  interior 
angles  between  the  parallels  A  C  and  BD,  and  are  therefore 
supplementary  (§  71). 

123.  Corollary  1.  All  the  angles  of  a  parallelogram  may 
he  determined  when  one  is  given,  the  angle  opposite  to  the 
given  one  being  equal  to  it,  and  the  other  two  angles  each 
equal  to  its  supplement. 


PARALLELS  AND  PARALLELOGUAm. 


55 


-M 


124.  Corollary  %,  1/ ttvo  parallelograms  have  one  angle 
of  the  one  equal  }n  one  angle  of  the  other,  all  the  remaining 
angles  of  the  one  will  be  equal  to  the  corresponding  angles  of 
the  other. 

125.  Corollary  Z.  If  one  angle  of  a  parallelogram  is  a 
right  angle,  all  the  other  angles  are  right  angles. 

Theorem  XXVIII. 

126.  A  pair  of  parallel  straight  lines  intercept 
equal  lengths  of  parallel  transversals. 

Hypothesis.  AB  and  CD,  any  pair  of  parallel  straight 
lines  ;  MN,  MS,  parallel  transverfjals  crossing  them  at  the 
points  M,  iV,  E,  and  S. 

Conclusion.    MN  =  R8.        a  \M 
ME  =  NS. 

Proof.    Join  the  two  oppo- 
site points  E  and  JV  by  a  third  C—  ^ \o    j^ 

transversal    EJV^,    and   compare 
the  two  triangles  MJVE  and  ^ES. 

1.  MENS  is  a  parallelogram,  by  definition.    Therefore 

Angle  EMN  =  angle  ESJV^.  (§  ng) 

2.  Because  EJ^ ia  a  transversal  crossing  the  parallels  AB 
and  CD, 

Angle  MEN=  alternate  angle  ENS.  (8  71) 

Angle  MNE  =  alternate  angle  NES. 

3.  Because  the  two  triangles  have  the  side  EN  common 
and  the  adjacent  angles  equal,  they  are  identically  equal 
(§109).   Therefore 

Side  ES  (opp.  angle  N)  =  side  MN  (opp.  equal  angle  E). 
Side  ME  =  corresponding  side  NS.     Q.E.D. 

127.  Corollary  1.  The  opposite  sides  of  a  parallelogram 
are  equal.  ^ 

128.  Corollary  2.  The  diagonal  of  a  parallelogram 
divides  it  into  two  identically  equal  triangles. 

If  the  two  transversals  are  perpendicular  to  the  parallels 

■ —   -j-.'va  xv,xxQt.ixo    mil  uiuiiiiuru  me  aisiance  of  the 

parallels.     Hence 


56 


BOOK  n.    BETILINEAL  FIOURES. 


i! 


/ 


1!S9.  Corollary^.  Two  parallels  are  everywhere  equaUy 
distant 

!  Theorem  XXIX. 

130.  If  three  or  more  parallels  intercept  equal 
lengths  upon  a  transversal  crossing  them,  they  are 
equidistant. 

Hypothesis.  A  transversal  crossing  the  parallel  lines  a, 
h,  c,  d,  etc.,  at  the  respective  points  AjB,  0,  />,  etc.,  in  such 
wise  that 

AB==  BO  =  CD,  etc. 

Conclusion.     The  distance  between  any  two  neighboring 
parallels,  as  «,  b,  is  equal  to  the 
distance  between  any  other  two, 
as  c,  d. 

Proof.  From  two  or  more  of 
the  points  of  intersection  A,  B,  6', 
etc.,  drop  perpendiculars  upon  the 
neighboring  parallels  and  consider 
the  triangles  thus  formed. 

1.  Because  the  angles  A,  B,C,  __ 
etc.,  i.re  corresponding  angles  be- 
tween parallels. 

Angle  A  —  angle  B  =  angle  G,  etc.  (§  71) 

2.  The  angles  X,  Y,  S,  etc.,  are  equal  by  construction, 
because  they  are  all  right  angles. 

3.  AB  =  BG=^  CD,  etc.,  by  hypothesis. 

4.  Therefore  any  two  triangles,.as  ABX,  CDS,  have  the 
angles  and  one  side  of  the  one  equal  to  the  angles  and  one 
corresponding  side  of  another,  and  are  identically  equal. 

5.  Therefore  AX=BY=  CS,  and  the  parallels  are  equi- 
distant.    Q.E.D. 

Theorem  XXX. 

131.  Conversely,  Equidistant  parallels  intercept 
equal  lengths  upon  any  transversal. 


PARALLELS  AND  PARALLELOGRAMS. 


67 


Hypothesis.     The  lines  a,  by  c,  d,  etc.,  parallel  and  equi- 
distant; MNy  a  transversal  inter- 
secting them  at  the  points^. 
By  Cy  Dy  etc. 

Conclusion.  Any  one  inter- 
cepted length  on  the  transversal, 
as  ABy  equal  to  any  other  inter- 
cepted length,  as  CD. 

Proof.  Take  up  the  figure 
and  lay  it  down  again  in  such 
manner  that  the  point  A  shall 
fall  into  the  position  C,  and  the 
line  a  coincide  with  the  trace  c.     Tuen 

1.  Because  angle  A  =  angle  C, 

Transversal  M^=  its  own  trace. 

2.  Because  the  parallels  a,  b  are  equidistant  with  the 
parallels  c,  d. 

Point  B  =  position  D. 
4.  Therefore  AB  =  CD.    Q.E.D. 

Scholium.    This  proposition  may  also  be  proved  by  drop- 
ping perpendiculars,  as  in  Theorem  XXIX. 


i 


•M 


Theorem  XXXI. 

133.  If  three  or  more  parallels  are  crossed  hy 
two  transversals  and  intercept  equal  lengths  on  one, 
then — 

I.  The  parallels  will 
intercept  equal  lengths  on 
the  other  transversal. 

II,  The     intercepted 
part  of  each  parallel  will 
he  longer  or  shorter  than 
the  neighboring  intercept  ^ 
hy  the  same  amount. 

Hypothesis.  GP,  0^,  any  e 
two  transversals;  A,  By  Cy 
equidistant  points  on   OP;  P 


"  'P 


68 


BOOK  II.    RECTILINEAL  FIGUMES. 


a,  b,  c,  d,  Cj  etc.,  parallels  through  these  points,  meeting  OQ 
in  the  points  A^,  B',  C",  etc. 

Conclusions.  I.  A'B'  =  B'0'=  CD'  =  D'E\  etc. 

II.  BB'  -AA'=  CQ'~  BB'=  DD'-  CC\  etc. 

Proof  I.  1.  Because  the  parallels  a,  h,  c,  etc.,  intercept 
equal  intervals  on  OP,  they  are  equidistant  (§  130). 

2.  Because  the  parallels  are  equidistant  they  intercept 
equal  intervals  on  the  transversal  OQ.  Therefore  the  inter- 
cepted intervals  A'B',  B'C,  etc.,  are  equal  (§  131). 

II.  Through  A'  draw  a  line  parallel  to  AB,  meeting  BB' 
in  K,  and  through  B'  draw  another  line  parallel  to  BC,  meet- 
ing C, C  in  L;  then — 

3.  Because,  by  construction,  ^^'J^iTand  BB'GL  are  paral- 
lelograms, 

BK=AA',) 

CL  =  BB',  S  ^§  134:) 

4.  Because  5'^ 'X  and  C"J5'Z  are  corresponding  angles  be- 
tween the  parallels  ^'JTand  B'L^  they  are  equal. 

6.  Because  A'B'K  and  B'C'L  are  corresponding  angles 
between  the  parallels  BB'  and  CC",  they  are  equal. 

6.  Comparing  with  (3)  the  triangles  A' KB'  and  B'LC, 
have  one  side  and  the  adjacent  angles  of  the  one  equal  to  a 
corresponding  side  and  two  adjacent  angles  of  the  other. 
Therefore  they  are  identically  equal,  and 

B'K^C'L. 

7.  By  construction, 

BB'  -BK=  B'K.  ^ 

CC  -  CL  =  CL. 
Comparing  with  3  and  6, 

BB'-AA'  =  B'X=CC-BB',  etc.    Q.E.D. 

133.  Corollary  1.  The  amount  by  which  each  length 
exceeds  the  preceding  one  may  be  found  by  laying  off  from 
the  point  of  intersection  0,  on  the  line  OP,  a  length  equal 
to  AB,  and  drawing  a  parallel  to  the  lines  a,  b,  etc.  The 
length  of  this  parallel  between  the  sides  OP  and  OQ  will  be 
equal  to  the  difference  between  the  lengths  of  any  two  con- 
secutive parallels. 


PARALLELS  AND  PARALLELOGRAMS. 


m 


134.  Corollary  2.    If  the  points  A,  B,  (7,  etc.,  are  found 
by  measuring  off  equal  distances  from  tlie  point  0,  so  that 

OA  =  AB  =  BC,  etc., q^ 

we  shall  have 

BB'=  %AA', 

CO'  =  3  A  A', 

DD'  =  4:AA% 

etc.  etc. 

135.  Corollary  3.  If  through  the  middle  point  of  one 
side  of  a  triangle  a  parallel  be  drawn  to  a  second  side,  it  will 
bisect  the  third  side  and  will  be  half  as  long  as  the  second  side. 

For,  let  OCD  be  the  triangle;  A,  the  middle  point  of  OC, 
and  ABj'd,  line  parallel  to  CD,  meeting  OD  in  B.  Let  a  third 
parallel  pass  through  0.  Then  0,  AB,  and  CD  are  three 
parallels  intercepting  equal  lengths  upon  the  transversal  OC. 
Therefore  they  also  intercept  equal  lengths  on  OD. 

Moreover,  CD  exceeds  AB  2ib  much  as  ^^  exceeds  the 
intercepted  length  of  0.  But  this  length  is  nothing.  There- 
fore CD  =  "HAB. 

Corollary  4.  Because  through  A  only  one  parallel  to  CD 
can  be  drawn,  it  follows  that  if  B  be  the  middle  point  of  02>, 
AB  will  be  that  parallel.     Therefore: 

136.  Tlie  line  joining  the  middle  points  of  any  two  sides 
of  a  triangle  is  parallel  to  the  third  side. 

Theorem  XXXII. 

137.  7)^  the  opposite  sides  of  a  quadrilateral  are 
equal  to  each  other ^  it  is  a  parallelogram. 

Hypothesis.    A  BCD,  a 
quadrilateral,  in  which 

AB  =  CD. 

AC  =  BD. 
Conclusion.  The  figure 

AB(^T)  \a  a.  TiJi.rallolrtorpQTTi 

Proof.  Draw  the  diagonal  BC.     Then — 

1.  The  three  sides  of  the  triangle  ABC  are  by  construe- 


li 


60 


BOOK  11.    RECTILINEAL  FIGUEE8. 


tion  and  hypothesis  respectively  equal  to  those  of  the  triangle 
DBC. 

2.  Because  the  angles  BCD  and  GBA  are  opposite  the 
equal  sides  CA  and  BD, 

Angle  BCD  =  angle  CBA.  (§  HO) 

3.  But  these  angles  are  alternatb  angles  between  the  lines 
CD  and  ^^  on  each  side  of  the  transversal  CB.    Therefore 

CD  II  AB, 

4.  In  the  same  way  may  be  shown  AC  \\  BD. 
Therefore  ABCDisa.  parallelogram,  by  definition.  Q.E.D. 

.Theorem  XXXIII. 

138.  If  any  two  opposite  sides  of  a  quadrilateral 
are  equal  and  parallel^  it  is  a  parallelogram. 

Hypothesis.   ABCDj  a  c_ ——.D 

quadrilateral  in  which 
CD  =  and  ||  AB. 

Conclusion.  AC=  and        / 
II  BD,  and  therefore  / 

ABCD  is  a  parallelogram.  -^ 

Prof.  Draw  the  diagonal  BC.     Then— 

1.  Because  AB  and  CD  are  parallels, and  5 C is  a  transversal, 

Angle  ABC=:  alternate  angle  BCD, 

2.  In  the  triangles  ABC  and  BCD, 

AB  =  CD,  by  hypothesis; 
^C  is  common; 
and  (1)  the  angles  between  these  equal  sides  are  equal.   There- 
fore these  triangles  are  identically  equal,  and 

AC=  BD.     Q.E.D. 
Angle  ACB=  angle  CBD. 

3.  T'i  30  angles  ACB  and  CBD  being  alternate  angles 
between  j±  C  and  BD, 

ACWBD.     Q.E.D. 

Theorem  XXXIV. 

139.  If  two  parallelograms  agree  in  the  lengths 
of  their  sides  and  in  one  angle^  they  are  identically 
equal. 


PARALLBLB  AND  PARALLELOGRAMS.  Qi 

Hypothesis,  ABCD  and  HKMN,  paraUelograms  in  which 
A.I3  —  UK. 

AC  =  HM, 

Angled  =  angled. 

Conclusion.  ABCD 
is  identically  equal  to 
HKMN, 

Proof,  ApyljABCD 
to  HKMNy  turning  it  ^ 
over,  if  necessary,  in  such  manner  that  the  angle  A  shall  coin- 
cide with  the  equal  angle  H,  and  the  side  AB  with  the  equal 
side  HK.     Then —  ^ 

1.  Because  these  sides  are  equal, 

Point  ^  E  point  JT. 

2.  Because  angle  A  =  angle  If, 

AC=HM. 

3.  Because  ^C=^J!/; 

Point  C=  point  Jf. 

4.  Because  the  parallelograms  have  the  ande  G  =  an^lfi 
M  (§  121),  ^  ^ 

Side  CD  =  MN. 

5.  In  the  same  way  it  may  be  shown  that  every  side  and 
angle  of  the  one  will  coincide  with  a  corresponding  side  and 
angle  of  the  other.     Therefore 

The  two  parallelograms  are  identically  equal.     Q.E.D. 
Scholium  1.  A  more  simple  but  less  general  demonstration 
may  be  given  by  drawing  diagonals  between  any  equal  angles. 

140.  Scholium  2.  We  have  shown  that  a  triangle  ip  com- 
pletely determined  when  its  three  sides  are  given.  From  the 
preceding  propositions  it  follows  that  a  quadrilateral  k  not 
completely  determined  when  its  four  sides  are  given,  but  chat 
the  angles  may  change  to  any  extent  without  changing  the 
lengths  of  the  sides. 

Thus  the  parallelogram  in  the 
margin  may  be  made  to  assume  the 

SnnP.PHHIVO  -(nvrna   ahrwirn     Vjtt  i-\\n  Ar^i- 

ted  lines.    This  is  the  geometrical 
expression  of  the  well-known  fact  It 


iU/ 


g-ff'iii  ji,[(n  n  "jinnBaj 


BOOK  II.    RECTILINEAL  FIGURES. 

thafc  a  frame  of  four  pieces  is  not  rigid  unless  fastened  by  a 
diagonal  brace. 

Theorem  XXXV. 

141.   The  diagonals  of  a  parallelogram  bisect 
each  other. 

Hypothesis.    ABGDy  a 
parallelogram  of  which  the 
diagonals  intersect  at  0. 
Conclusions.  CO  =  OB. 
AO^OD. 
Proof.  In  ttf  triangles  A  OB  and  CODy 
Angle  AOB  =  opposite  angle  COD. 


7^ 


"^ 


X 


Angle  BAO  =  alternate  angle  ODC, ) 
Angle  ABO  =  alternate  angle  DCO.  ) 


(§67) 

(§  71) 
(§124) 


Side  ^5  =  side  CD. 
Therefore  the  two  triangles  are  identically  equal,  and  the 
sides  opposite  the  equal  angles  equal;  namely, 

BO  =  OC. 
AO=OD.    Q.E.D. 


MiaOELLANEOm  PROPEBTIES. 


68 


CHAPTER   IV. 

MISCELUNEOUS  PROPERTIES  OF  POLYGONS. 


Befinitions. 

.1.  ?^*^;  -P'S^-   A  polygon  is  a  figure  formed  by  a 
chain  of  straight  hnes  returning  into  itself  and  inclos- 
ing a  part  of  the  plane  on  which  it  Hes. 
The  lines  are  called  sides  of  the  polygon 


Polygons  of  elementarj  geometry. 

..l""  !^'  ^'tT  j^^^°^^<^^y  *^«  «ide8  of  a  polygon  may  cross 
each  other  but  elementary  geometry  treats  only  of  polygons 
the  sides  of  which  dc  not  cross.  ^^ 

143.   The  angles  of  the  polygon  measured  on 
the   side   containing   the   in- 
closed space  are  called  inte- 
rior  angles. 

Example.  One  interior  angle 
of  the  polygon  ABODE  is  the 
angle  EAB,  measured  by  turning 
the  side  AB  through  the  interior 
of  the  polygon  until  it  coincides 
with  AE.     In  the  first  polygon 


the 


ontAaa 

— o 


01«A 


txi.z 


xcso 


Polygon  the  sides  of  which  cross. 


,  u .wx.^  ^M^wm^^        -  •"■J  evu  vuv  otucB  ML  WIUCU  CFOSS. 

Straight  angles.    But  in  the  polygon  ABODE  the  interior 
angle  AED  is  greater  than  a  straight  angle 


64 


BOOK  II.    RECTILINEAL  FIGURES. 


144.  An  exterior  angle  of  a  polygon  is  an  angle 
between  one  side  and  the  continuation  of  the  adjacent 
side.  '' 

Eemark  1.  To  form  all  the  different  exterior  angles  of  a 
polygon  It  18  sufficient  to  produce  each  side,  taken  in  regular 
order,  in  one  direction.  The  number  of  exterior  angles  will 
then  be  the  same  as  that  of  the  interior  angles. 

The  angles  BAL,  GBO,  etc.,  are  exterior  angles     Also 
Angle  ABC-{-  angle  CBQ  =  straight  angle.  ' 


The  angle  PQR  is  an  exterior  angle  which  falls  inside  the  polygon 
because  the  interior  angle  PQS  is  greater  than  a  straight  angle. 

Remark  2.  It  is  evident  that  a  polygon  has  as  many 
angles  as  sides. 

Remark  3.  If  the  interior  angle  of  a  polygon  is  greater 
than  a  straight  angle,  the  continuation  of  one  of  the  sides  will 
fall  within  the  polygon. 

The  corresponding  exterior  angle  is  then  regarded  alge- 
braically as  having  the  negative  sign. 

Remark  4.  It  is  evident  that  the  sum  of  any  interior 
angle  and  of  the  corresponding  exterior  angle  is  a  straight 
angle;  that  is. 

Int.  angle  +  ext.  angle  =  180°. 
By  transposing  the  second  term  of  the  first  members,  we  have 

Ext.  angle  =  180°  -  int.  angle. 
If  the  interior  angle  is  greater  than  180°,  the  members  of 
this  eo^jiiition  will  be  negative. 


MISCELLANEOUa  PBOPERTlEa.  55 

Classification  of  Polygons. 

146.  Polygons  are  classified  according  to  the  num- 
'  ber  of  their  sides. 

The  least  number  of  sides  which  a  polygon  can  have 
is  three,  and  it  is  then  a  triangle. 

146.  Def.  A  quadrilateral  is  a  polygon  of  four 
sides. 

147.  Def.   A  pentagon  is  a  polygon  of  five  sides. 

148.  Def,  A  hexagon  is  a  polygon  of  six  sides. 

149.  Def  An  octagon  is  a  polygon  of  eight  sides. 

150.  Polygons  of  any  number  of  sides  may  be 
designated  hj  the  Greek  numerals  expressing  the 
number  of  sides.  ^ 

151.  Def  A  diagonal  of  a  polygon  is  a  Kne  join- 
ing any  two  non-adjacent  angles. 

Question  for  the  student.  How  many  diagonals  can  be 
drawn  from  one  angle  of  a  polygon  having  n  sides  ? 

152.  A  regular  polygon  is  one  of  which  the  sides 
and  angles  are  all  equal. 

Classification  of  Quadrilaterals. 

16  .  Def  A  trapezoid  is  a  quadrilateral  of  which 
two  opposite  sides  are  parallel. 

154.  If  both  pairs  of  opposite  sides  are  parallel, 
the  quadrilateral  is  a  parallelogram. 

155.  i>e/.  A  rectangle  is  a  parallelogram  in  which 
the  four  angles  are  equal. 

XT.  \^^'  Py-  ^  ^^ombus  is  a  parallelogram  of  which 
the  lour  sides  are  equal. 


*  -I 


sides  are  equal 


Def  A  square  is  a  rectangle  of  which  the 


66 


BOOK  11.    RECTILINEAL  FIQURES. 


— f» 


Theorem  XXXVI. 

158.  JJ^  each  side  of  a  polygon  he  produced  in 
one  direction^  the  sa.r  o/  all  the  exterior  angles  is 
equal  to  a  circuiiifereiuie. 

Hypothesis.  ABCDE,  a 
polygon. 

Conclusion.  Sum  of  n  ex- 
terior angles  ABC,  etc.  =  two 
straight  angles. 

Proof.  From  any  point  0 
draw  n  straight  lines,  each 
parallel  to  one  of  the  sides  of 
the  polygon.  There  will  then 
be  n  angles  around  the  point  0. 

1.  Because  OP  is  parallel  to  BK  and  similarly  directed, 
and  OQ  io  BC  and  similarly  directed, 

Angle  POQ  =  exterior  angle  KBO.  (§  79) 

2.  In  the  same  way  it  may  be  proved  that  each  of  the 
exterior  angles  of  the  polygon  is  equal  to  one  of  the  angles 
around  0. 

3.  Because  the  number  of  exterior  angles  and  of  angles 
around  0  is  equal,  the  sum  of  all  the  exterior  angles  will  be 
equal  to  the  sum  of  all  the  angles  around  0;  that  is,  to  a 
circumference.     Q.  E.  D. 

159.  Scholium.  If  any  of  the  interior  angles  of  the 
polygon  should  be  reflex,  so  that  exterior  angles  fall  within 
the  polygon,  such  exterior  angles  must  be  regarded  as  alge- 
braically negative  in  forming  the  sum  (§  144). 


Theorem  XXXVIL 

160.  The  sum  of  the  interior  angles  of  a  polygon 
is  equal  to  a  number  of  straight  angles  two  less  than 
the  number  of  sides  of  the  polygon. 

Hypothesis.  ABCDEF,  a  polygon  having  n  sides  and  n 
angles. 


MISCELLANEOUS  PROPERTIES. 


67 


-P 


K 


Conclusion,   Sum  of  the  n  interior  angles  equal  to  n  —  2 
straight  angles,  or 
ABC  +  BCD  -f  CDE  +  etc.  = 
(w-2)180°. 

Proof.  1.  The  sum  of  each  in- 
terior angle  and  it's  adjacent  exterior 
angles,  as  ABO-\-CBOy  is  a  straight 
angle  (§  61). 

3.  Because  the  polygon  has  n 
such  pairs  of  angles,  the  sum  of  all 
the  interior  and  exterior  angles  is  n 
straight  angles.  Xi 

3.  But  the  sum  of  the  exterior  angles  alone  is  a  circnm- 
ference;  that  is,  two  straight  angles  (§  155). 

4.  Taking  these  two  straight  angles  from  the  sum  (2),  we 
have  left  the  sum  of  the  interior  angles  aloue^  equal  to  n  —  2 
straight  angles.    Q.E.D. 

161.  Corollary  1.  The  sum  of  all  the  interior  angles 
of  a  quadrilateral  is  equal  to  two  straight  angles;  that  is,  to 
four  right  angles. 

163.  Corollary  2.  Since  in  a  rectangle  all  the  four 
angles  are  equal,  and  their  sum  is  four  right  angles,  each  of 
the  angles  is  a  right  angle. 

Theorem  XXXVIII. 
163.  If  through  each  angle  of  a  triangle  a  line 
he  drawn  parallel  to  the  opposite  side.,  the  three  lines 
will  form  a  triangle  the  sides  of  which  will  he  hisected 
hy  the   ertices  of  the  original  triangle. 

Hypothesis.  ABC,  any  tri-  B^ 
angle;  DEF,  another  triangle 
formed  by  drawing, 
through  Cy  J^Z)parallelto^5; 
through  Ay  -fi^i^parallel  to  CB\ 
through  By  FD  parallel  to  A  C, 
Conclusion,    EC—  CD. 
EA  =  AF-. 
FB  =  BD. 


im 


es 


BOOK  IL    RECTILINEAL  FIGUUES. 


Proof.    The  quadrilaterals  ABCD  and  ABEC  are  paral- 
lelograms, by  construction. 
Therefore 

CD  =  ABi ) 

EC=:AB.\  (§127) 

Whence 

EC=CD.    Q.E.D. 
In  the  same  way  the  other  conclusionfl  may  be  preyed. 

Theorem  XXXIX. 

164.  The  bisectors  of  the  three  interior  angles  of 
a  triangle  meet  in  a  point  equally  distant  frmi  the 
sides  of  the  triangle. 

Hypothesis.  ABC,  any  triangle; 
AO,  BO,  COy  the  bisectors  of  its  in- 
terior angles,  A,  B,  and  C. 

Conclusions.  I.  These  three  bisect- 
ors meet  in  a  single  point  0. 

II.  This  point  0  is  equally  distant 
from  the  three  sides  of  the  triangle. 

Proof.  Let  0  be  the  point  in  which 
the  bisectors  of  ^  and  5  meet.  Then— 

1.  Because  0  is  on  the  bisector  of  the  angle  A,  O'ls  equally 
distant  from  the  sides  AB  and  .4Cof  the  angle  A  (§  106). 

2.  Because  0  is  on  the  bisector  of  the  angle  B,   0  is 
equally  distant  from  the  sides  BA  and  ^6' of  the  angle '5. 

3.  Therefore  0  is  equally  distant  from  ^(7  and  BC. 

4.  Therefore  it  is  upon  the  bisector  of  the  angle  formed 
by  ^Cand  BC]  namely,  of  the  angle  C  (§  107). 

5.  Therefore  the  point  0  is  equally  distant  from  the  three 
sides,  and  the  bisectors  all  pass  through  it.     Q.E.D. 

Theoeem  XL. 

165.  The  bisectors  of  any  two  exterior  angles  of 
a  general  triangle  meet  the  bisector  of  the  opposite 
interior  angle  in  a  point  which  is  equally  distant 
from  the  three  sides  of  the  triangle. 


MISOELLANEOUa  PROPERTIES. 

Hypothesis.    ABC,  any  triangle,  of  whicli  the  sides  AB 
and  AC  are   produced  indefinitely  in  the  directions  P  and 
Q'y   BO,   CO,  the  bisectors  of 
the  exterior  angles  CBP  and 
BCQ;  0,  the  point  of  meeting 
of  these  bisectors. 

Conclusions.  I.  The  bisector 
of  the  angle  BAC  passes 
through  0. 

II.  The  point  0  is  equally 
distant  from  the  lines  BC,  BP, 
and  CQ. 

Proof,    1.  Because  0  is  on  > 
the  bisector  of  the  angle  CBP,  '^ 
it  is  equally  distant  from  the 
sides  -5(7  and  BP. 

2.  Because  0  is  on  the  bisector  of  the  angle  BCQ,  it 
is  equally  distant  from  the  sides  ^Cand  CQ. 

3.  Therefore  0  is  equally  distant  from  the  three  lines  AP, 
BC,&TidAQ.    Q.E.D. 

4.  Because  0  is  equally  distant  from  AP  and  AQ,  it  is  on 
the  bisector  of  the  angle  made  by  those  lines  (§  107). 

Therefore  this  bisector  passes  through  0.     Q.E.D. 


Theorem  XLI. 

166.  The  perpendicular  bisectors  of  the  three 
sides  of  a  triangle  meet  in  a  point,  which  point  is 
equally  distant  from  the  three  vertices  of  the  triangle. 

Hypothesis.  ABC,  any  triangle;  R, 
B,  Q,  the  middle  points  of  the  respective 
sides;  PO,  BO,  lines  passing  through  P 
and  R  perpendicular  to  BC  and  AB 
respectively. 

Conclusions.  I.  The  point  0  is  equally 
distant  from  A,  B,  and  C. 

il.  The  perpendicular  bisector  ot  AC  A 
passes  through  0. 


\m^\m 


70 


BOOK  II.    RECTILINEAL  FIGURES. 


Proof.  1.  Because  0  is  on  the  perpendicular  bisector  of 
the  line  BCy  it  is  equally  distant  from  the  ends,  5  and  C  of 
this  line  (§  104). 

2.  Because  0  is  on  the  perpendicular  bisector  of  the  line 
ABy  it  is  equally  distant  from  the  points  A  and  B. 

3.  Therefore  it  is  equally  distant  from  the  three  points  A 
By  and  G.     Q.E.D.  ^ 

4.  Because  it  is  equally  distant  from  A  and  C,  it  lies  on 
the  perpendicular  bisector  of  the  line  JC(§  105). 

Therefore  this  bisector  also  passes  through  0.    Q.E.D. 


Theorem  XLIL 

16*7.  The  perpendiculars  dropped  from  the  three 
angles  of  a  triangle  upon  the  opposite  sides  pass 
through  a  point. 

Hypothesis.  A  BCy  any  triangle;  AQ,  BR,  CP,  perpen- 
diculars from  A,  By  C,  upon  BCy  CA,  AB,  respectively. 

Conclusion.     These  perpendiculars  pass  through  a  point. 

Proof.    Through  Ay  B,  and  (7,  respectively,  draw  parallels 
to  the  opposite  sides  of  the  trian- 
gle, forming  the  triangle   LMJ^.  \, 
Then—  •  ^ 

1.  Ay  B,  and  (7  will  be  the  mid- 
dle point  of  the  sides  of  the  trian- 
gle LMN  (§  163). 

2.  Because  LM  is  parallel  to 
ABy  and  CP  perpendicular  to  AB, 
CP  is  also  perpendicular  to  LM. 

(§72) 

3.  Therefore  CP  is  the  perpen- 
dicular bisector  of  LM.  In  the 
same  way  BR  and  A  Q  are  perpen- 
dicular bisectors  of  LN  and  MN 
respectively. 

4.  Because  the  three  lines  A  Qy 
BRy  and  CP  are  the  perpendicular  bisectors  of  the  sides  of 
the  triangle  LMJVy  thoy  pass  through  a  point  (§  166).     Q.  E.  D. 


MiaCELLANBOUS  PB0PEBTIE8. 


71 


168.  Def.  The  line  drawn  from  any  angle  of  a  tri- 
angle to  the  middle  point  of  the  opposite  side  is  called 
a  medial  line  of  the  triangle. 

Corollary.  Since  a  triangle  has  three  angles  it  may  have 
three  medial  lines. 


>' 


Theorem  XLIII. 

169.  The  three  medial  lines  of  a  triangle  meet  in 
a  point  which  is  two  thirds  of  the  way  from  each 
angle  to  the  middle  of  the  opposite  side. 

Hypothesis.  ABC,  sl  triangle;  F,  Q,  R,  the  middle  points 
of  its  respective  sides; 
BE,  AQ,  two  medial 
lines  of  the  triangle;  0, 
their  point  of  intersec- 
tion. 

Conclusions.  I.  The 
third  medial  line  CF 
also  passes  through  the 
point  0. 


II. 


\  QO  =.  ^OA. 
( RO  =  iOB. 


Froof.  Bisect  AO  in  M,  and  OB  in  JV.    Join  RQ,  RM. 
QN,MN.     Then— 

1.  Because  MN  is  a  line  joining  the  middle  points  of  the 
sides  OA  and  OB  of  the  triangle  OAB, 

MN=^AB.  (§135) 

MN  II  AB.  (§  136) 

2.  Because  i?^  is  a  line  joining  the  middle  pomts  of  the 
sides  CA  and  CB  of  the  triangle  ABC, 

RQ  =  {AB, 
RQ  II  AB, 

3.  Therefore  RQh  parallel  and  equal  to  MN,  whence  the 
quadrilateral  RQMN  is  a  parallelogram  (§  138). 


^ 


?MmmmM.mmmmi^jmmmm 


72 


BOOK  II.    RECTILINEAL  FIGURES. 


RQMn''^''^^  ^^^^  ^^"^^  diagonals  of  the  paraUelogram 

;  '  OQ=OM.l 

OH  =OJV.\  (§  1^1) 

But,  by  construction,  M  and  TiT  are  the  bisectors  of  OA  and 
Olf,    Therefore 

OM=iOA. 

Whence  ^^=^^^- 

QO=:iOA, 

JiO  =  iOB.     Q.E.D. 
^1,,-i  ^1  *^®  same  way  it  may  be  shown  that  th«  point  in 
which  the  medial  line  CP  cuts  the  medial  line  ^^  is  two 

The  three  medial  lines  pass  through  the  point  0,   Q.E.D. 
Bjl'tllr^  '''''  ^^  '^'  same  way  as  withi^Q  and 

PO  =  iOa    Q.E.D. 

Theorem  XLIV. 
170.  '^^e  line  drawn  from  the  middle  of  one  of 

Z.^^^;^  ^  /.a?/  ^;^.^r  sum,  and  bisects  the  opposite  side 

theSlfrsidelTt^^^^^^^^^     ''  ^^^^^  -^  '^  - 
middle  point  of  A  C\  EF, 

a  parallel  to  AB  and  CD, 

meeting  BD  in  F. 

•    Conclusions. 

I.  EF  is  equidistant 
from  AB  and  Ci>. 
11.  BF=zFD. 
lll-EF=^AB-^CD). 

cJrth  ?ZT  A^.  f^t -^  ^^  are  parallels  inter- 
they  a^^disS^C^^r^^r^  '''  *""""^^  ^^ 


MiaCELLANEOUa  PROPEHTIES.  73 

2.  Because  they  are  equidistant,  they  intercept  equal 
lengths  upon  the  transversal  BD  (§  131).    Hence 

BF^FD.    Q.E.D. 

3.  Therefore  EF  -  CD  =  AB  -  EF  {%  132,  II.),  and  by 
transposition 

%EF=AB-ir  CD, 

c»  EF=i(AB+CD).     Q.E.D. 

Def.  The  line  ^i^'is  called  the  middle  parallel  of 
the  trapezoid. 

Theorem  XLV. 

1^:11.  If  the  two  non-parallel  sides  of  a  trapezoid 
are  equal,  the  angles  they  make  with  the  parallel  sides 
are  equal. 

Hypothesis.  ABCD,  a  trapezoid  in  which  AB  and  67)  are 
parallel,  and  CA  =  DB. 

Conclusion. 
Angle  CAB  =  angle  DBA. 
Angle  ACD  =  angle  BDC. 

Proof.  From  C  draw  CE 
parallel  to  DB,  meeting  AB 
in  E.     Then— 

i.  Because  CE  is  parallel  to  DB,  and  CD  to  EB, 
DBCE  is  a  parallelogram. 

2.  Because  D^C^  is  a  parallelogram, 

CE=DB. 

3.  Because  by  hypothesis  DB  =  CA, 
Therefore  CE  =  CA. 

4.  Therefore  A  CE  is  ar*  isosceles  triangle,  and 

Angle  CAE=  angle  CEA. 
6.  Because  CE  and  /)j5  are  parallel. 

Angle  DBE  =  corresponding  angle  CEA, 
6.  Comparing  with  (4), 

Angle  CAE  =  angle  DBE.     Q.E.D. 
In  a  similar  way,  by  drawing  through  A  a  parallel  to  BD,  is 


it 


H 


oh  it   Twi 


Angle  ^  Ci>  =  angle  i?i)  C     Q.  E.  D. 


fflglWaB'^l^RaawaBiiM 


74 


BOOK  II.    RECTILINEAL  FIGURES. 


B 


Theorem  XLVI. 
17:3.  Conversely,  if  the  angles  at  the  base  of  a 
trapezoid  are  equal,  the  non-parallel  sides  and  the 
other  angles  are  equal. 

Hypothesis.   ABCD,  a  trapezoid  in  which  the  sides  AB 
and    CD  are    parallel   and 
Angle  CAB  =  angle  DBA. 

Conclusion. 
I.  CA  =  DB. 

II.  Angle  ^C7)=angle5Z)a 

Proof.  Make  the  same 
construction  as  in  the  last 
theorem.     Then — 

1.  Because  BA  is  a  transversal  crossing  the  parallels  CE 
and  DB,  Angl6  CEA  =  angle  DBE. 

2.  By  hypothesis, 

Angle  DBE  =  angle  CAE. 
Whence  Angle  CAE  =  angle  CEA. 

3.  Therefore  A  CE,  having  the  angles  at  the  base  equal,  is 
an  isosceles  triangle,  and 

CA=^CE=zDB.     Q.E.D. 
Therefore  angle  ACD  =  angle  BDG  (§  171).     Q.E.D. 

Eemark.   This  form  of  trapezoid  is  sometimes  called  an 
antiparallelogram. 

Theorem  XLVII. 
173.^  quadrilateral  of  which  two  adjoining  pairs 
of  sides  are  equal  is  symmetrical 
with  respect  to  the  diagonal  Join- 
ing the  angles  formed  by  the 
equal  sides,  and  the  diagonals 
cut  each  otJver  at  right  angles. 

Hypothesis.  A  B  CD,  a  quadrilateral 
in  which        AB  =  AD. 
CB  =  CD. 

Conclusion.   ACis  &n  axis  of  sym- 
metrv.     A  (^  cuts  BD  at  riffht  ancrles. 

Proof,    Because,  in  the  triangles 
ABC  m^  ADC, 


MISCELLANEOUS  PEOPEMTJES. 


76 


ACi&  common, 
these  triangles  are  identically  equal,  and 

Angle  BA  G  (opp.  BC)  =  angle  DA  0  (opp.  A  C). 
Angle  ^(7^  (opp.  AB)  =  angle  DGA  (opp.  AD), 
Therefore,  if  the  figure  be  turned  over  on  the  line  ^f— 
The  lines  AB  and  CB  will  fall  upon  ADwidiCD  respectively. 
The  point  J5  «         «    the  point  i). 

And^i)  «         «    its  own  trace. 

Therefore  the  figure  is  symmetrical  and  the  line  BD  at 
right  angles  to  A  G.    Q.  E.  D. 

Lemma  RESPECTijjq^G  Identical  Figures. 

174.  In  identically  equal  figures,  corresponding  lines  are 
equal. 

Note.  Corresponding  lines  are  those  which  coincide  when  the 
figures  are  applied  to  each  other.  From  this  definition  the  conclusion 
follows  without  demonstration. 

Corollary,  In  identical  figures,  any  lines  so  defined  that 
there  can  be  but  one  line  in  each  figure  answering  to  the 
definition  are  corresponding  lines. 

For  if  such  lines  did  not  coincide  when  the  figures  were 
brought  into  coincidence,  the  two  lines  would  equally  cor- 
respond to  the  definition. 

175.  Special appUcaiions.  In  identically  equal  triangles— 

The  perpendiculars  from  equal  angles  upon  the  opposite  sides 
are  equal. 

The  Usec*ors  are  of  equal  length. 

In  ^  c  JA>.ally  equal  quadrilaterals  the  diagonals  are  equal. 


its 
t. 


76 


BOOK  II    RECTILINEAL  FIQURE8. 


CHAPTER   V. 

PROBLEMS. 


Postulates. 
176.  It  is  assumed — 

inflpfi*  .Tf'i^*-''  ^.'?'*^  f*'""^^^*  ^^^^  "^^y  be  produced 
indetL^ite]y  m  eitlier  direction. 

III.  That  a  circle  may  be  drawn  around  any  point 

a  n?ipv  ''^;^""^^^^*«  f  *^««^  P^st^l'^tes  may  be  fulfilled  with 
a  ruler  and  a  pair  of  compasses,  which  are  the  only  instru- 
ments recognized  m  pure  geometry. 

But  it  is  not  necessary  to  confine  one»s  self  to  these  instru- 
ments m  solying  all  problems.  When  it  is  once  well  under- 
stood how  a  given  probJem  is  to  be  solved  by  them,  other 
instruments  may  be  used  for  the  actual  drawing,  such  as  the 
protractor,  the  square,  and  the  parallel  ruler. 

Peoblem  I. 

177.  Onagirm  straight  line  to  marJc  off  a  lenath 

equal  to  a  gimnflnite  straight  line. 

Given   AB,  an  indefinite  straight  line;  «,  a  given  finite 
straight  line.  ,  * 

Required.   To  mark  off               ; 
on  J^  a  length  equal  to  a.    ^' ^] ~ B 

Construction.  From  any  \ 

point  0  as  a  centre,  on  AB  « 

describe  the  arc  of  a  circle  with  a  radius  equal  to  «.  K  p  be 
the  point  m  which  the  circle  intersects  a\  OP  will  be  the 
required  length.  wm  oe  me 

Proof  AH  the  radii  of  the  circle  around  0  are  b  v  construe 
tion  equal  to  ..  OP  is  one  of  these  radii;  therefore  itlqual  L  ^^ 


PROBLEMS. 


77 


g 


Problem  II. 

178.  To  construct  a  triangle  of  wMcTi  the  sides 
shall  he  equal  to  three  given  straight  lines. 

Given.  The  three  lines 
a,  d,  c. 

Required.  To  draw  a 
triangle  with  sides  equal  to 
these  lines. 

Construction.  On  an  in- 
definite line  take  a  length 

CB  equal  to  any  one  of  the         '  . 

three  given  lines  (Problem  

I.), 

From  B  as  a  centre,  with  a  radius  equal  to  one  of  the 
remaining  lines,  c,  describe  the  arc  of  a  circle. 

From  the  point  C  as  a  centre,  with  a  radius  equal  to  the 
third  given  line  describe  another  arc  of  a  circle  intersecting 
the  former  one  in  a  point  A.     Join  CA  and  BA. 

ABC  will  then  be  the  triangle  required. 

Proof.  From  the  mode  of  construction,  the  three  lines 
AB,  AC,  and  BC  will  be  equal  to  the  three  given  lines. 

Kemark.  The  two  circles  may  intersect  on  either  side  of 
the  line  AB.  Therefore  tv;o  triangles  may  be  drawn  which 
shall  fulfill  the  given  conditions.  But  these  triangles  will,  by 
§  110,  be  identically  equal. 

Problem  III. 
179.  To  bisect  a  given  finite  straight  line. 

Given.    The  line  AB. 

Required.     To    bisect    it. 

Construction.      From    the 
end  ^   as  a  centre,   with    a 
radius  greater  than  the  half  of  . ,  -'' 
AB,  draw  the  arc  of  a  circle 
CND. 

From  J5  as  a  centre,  with 

■r\\  r\       actmrt  r\       -»«rt /l  i  ii  o       r\-nr%-%tT     4-T%  ^      ^-^-h^ 

CMD,   intersecting    the    first 

circle  in  the  points  Cand  D.      Join  CD. 


/'^^^. 


mT 


'N 


^B 


/y 


/ 


^ 


78 


BOOK  11.    BEOTILINEAL  FIGUBB8. 


The  point  0  in  which  the  line  CD  intersects  AB  will  then 
bisect  AB. 

Proof.     Join  AC,  BC,  AD,  BD.  ^ 

Because  AC,  AD,  BC,  BD,  are  radii  of  the  same  or  equal 
circles,  they  are  equal,  and  the  figure  ADBC  is  a  paral- 
lelogram (§134). 

Therefore  the  diagonals  AB  and  CD  bisect  each  other. 

(§  138) 

Therefore  ^J5  is  bisected  at  0.     Q.E.F. 

180.  Corollary.  Because  the  parallelogrom  A  BCD  has 
all  its  sides  equal,  it  is  a  rhombus,  and  its  diagonals  intersect 
at  right  angles.  Therefore  the  abovo  construction  also  solves 
the  problem: 

To  draw  the  perpendicular  bisector  of  a  given  line. 


Problem  IY. 

181.  To  bisect  a  given  angle. 

Let  A  CB  be  the  given  angle. 

Construction.  From  C  as  a  centre,  with  any  radius  CA 
describe  the  arc  of  a  circle,  cutting  the  sides  of  the  angles  in 
the  points  A  and  B. 

From  ^  as  a  centre,  with  the  radius  AB  draw  an  arc  of  a 
circle. 

From  5  as  a  centre,  with  an  equal  radius  draw  another  arc 
intersecting  the  other  in  0. 

Join  CO. 

The  line  CO  will  bisect  the  given  angle  A  CB, 

Proof.     In   the  triangles 
C^Oand  C^Owehave 

„  .  ~  ^.  „'  !■  by  construction. 
OA  =  OB,  )    ^  c 

CO  =  CO,  identically. 

Therefore  these  two  trian- 
gles are  identically  equal,  and 
the  angle  OCA,  opposite  the  side  AO,  is  equal  to  OCB,  oppo- 
site the  equal  side  in  the  other  triangle.  Therefore  the  line 
CO  bisects  the  angle  A  CB.     Q.  E.  F. 


PROBLEMS. 


Problem  V. 


79 


k 


B' 


182.  Through  a  given  point  on  a  straight  line  to 
draw  a  perpendicular  to  this  line. 

Let  MN  be  the  given  line; 
Of  the  given  point  upon  it. 

Construction.  From  0  as  a 
centre,  with  any  radius  OA  de- 
scribe arcs  of  a  circle  cutting 
the  given  line  at  A  and  B. 

From  ^  as  a  centre,  with 
the  radius  AB  draw  the  arc  of  ..  / 
a  circle.  SI4A. 

From  5  as  a  centre,  with  the  equal  radius  BA  describe 
another  arc  intersecting  the  former  one  at  O.    Join  OG. 

OG  will  be  the  required  perpendicular. 

Proof.  In  the  triangles  CUO  and  GBO  the  three  sides  of 
the  one  are,  by  construction,  equal  to  the  three  sides  of  the 
other. 

Therefore  the  angle  GAG  ib  equal  to  the  angle  GOB,  and 
both  these  angles  are  right  angles,  by  definition. 

Therefore  the  line  OGi&  perpendicular  to  MN.    Q.E.F. 

Problem  VI. 

183.  From  a  given  point  without  a  given  line  to 
drop  a  perpendicular  upon  the  line. 

Let  P  be  the  given  point; 
MNj  the  given  line. 

Gonstruction.  Take  any 
point  K  on  the  opposite  side  of 
the  line.  From  P  as  a  centre, 
with  the  radius  PK  describe 
an  arc  cutting  the  given  line  at 
A  and  B. 

Bisect  AB  in  the  point  0.    Join  PO, 

Tl\e  line  PO  will  be  the  perpendicular  required. 


^     r  vxrj  E 


xxo 


student. 


in  viiO  xaa\j  problem,  und  to  be  supplied  by  the 


II 


80 


BOOK  II.    RECTILINEAL  FIOUliES. 


PltOBLEM  VII. 

184.  At  a  given  point  in  a  straight  line  to  make 
an  angle  equal  to  a  given  angle. 


Given.  Anmg\Q,EFG'y  a  straight  line,  J ^;  a  point  0  on 
that  line. 

Required.  At  0  to  make  an  angle  equal  to  EFG. 
Construction.    1.  Join  any  two  points  in  the  sides  FE  and 
FG,  thus  forming  a  triangle. 

2.  FromO  take  on  OB  a  distance  0K=-  FE. 

3.  On  0^  describe  a  triangle  OJOf  whose  sides  KM,  OM 
shall  be  equal  to  the  sides  EG,  FG. 

4.  The  angle  MOK  will  be  the  required  angle. 

Proof.  Because  the  triangles  OiOf  and  ^jE'G^  have  all  the 
sides  of  the  one  equal  to  corresponding  sides  of  the  other,  they 
are  identically  equal. 

Therefore  the  angle  MOK,  opposite  MK,  is  equal  to  the 
angle  EFG,  opposite  the  equal  side  EG  of  the  other  triangle. 
Q.E.F. 

Problem  VIII. 

185.  To  construct  a  triangle,  having  given  two 
ndes  and  the  included  angle. 

Given.    Two  sides,  «,  J;  the  angle  ^. 

Required.  To  construct 
a  triangle  having  the  sides  Q^ 

equal   to  a,  b,  and  their 
included  angle  equal  to  g. 

Construction.  Draw  an 
indefinite  line  AB. 

At  A  make  the  angle 
BA  C  equal  to  g. 

On  AB  take  a  length 


Q-nH    r\tr\ 


A  n  f  oi-.- 


uii  ^-xi>  i/iiKu  a 


length  A  Q  equal  to  a. 


PROBLEMS,  g| 

Join  PQ. 

APQ  will  then  be  the  required  triangle. 

The  result  is  evident,  but  should  be  shown  by  the  pupil. 

Problem  IX. 

186.  Two  angles  of  a  triangle  being  given,  to  ccm- 
strw  '  the  third  angle. 

Given.  Two  angles,  c  and  e,  of  a  triangle. 

Required.  To  find  the 
third  angle  of  the  triangle. 

Constructmi.  1.  At  any 
point  0  in  an  indefinite  line 
AB  make  the  angle  BOG 
equal  to  the  given  angle  c.  \  ^     ^C 

2.  At  the  same  point  make 
the  angle  COD  =  given  an- 
gle e.  -a. ^ B 

Then  the  angle  DO  A  will  be  the  angle  required. 

Proof.  From  Theorem  IV.,  to  be  supplied  by  the  student. 

Problem  X. 

187.  To  construct  a  triangle,  having  given  one 
side  and  the  two  adjacent  angles. 

Given.  A  finite  straight ^ 

line,  AB;   two  angles,  c     ^ 
and  e. 

Required.  To  construct 
a  triangle  having  its  base 
equal  to  ABf  and  the  an- 
gles at  its  base  equal  to  c 
and  e  respectively. 

Construction.  1.  On  an 
indefinite  straight  line 
mark  off  the  length  AB. 

2.  At  A  make  the  angle  BAM  =  angle  c, 

3.  At  B  make  the  angle  ABN=z  angle  e. 

4.  Continue  the  lines  AM  and  Bly  until  they  meet,  and 
let  C  be  their  point  of  meeting. 


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ABOmH  then  be  the  required  triangle. 

The  proof  should  be  supplied  by  the  student. 

OoroUary.  Since  the  third  angle  of  a  triangle  can  always 
be  found  when  two  angles  are  giyen,  this  problem,  combined 
with  tne  precedmg  one,  will  suffice  for  the  construction  of  the 
triangle  when  one  side  and  any  two  angles  are  giyen. 

Pboblem  XI. 

188.  Two  sides  of  a  triangle  and  the  angle  appo- 
site one  of  them  being  given,  to  construct  the  triangle. 

Given,  The  two  sides,  «— ■ 

«,  h;  the  angle  B  opposite 
the  side  h. 

Required,  To  con- 
struct the  triangle. 

Construction.  1.  Ai! 
the  point  B  on  the  indefi- 
nite line  BM  make  an 
angle  MBR  equal  to  the 
given  angle  B. 

2.  From  ^  on  the  line       B  U"^^^ — — '-jf 
BR  cut  off  a  length  ^C  equal  to  the  giyen  line  «,  which  is 
not  opposite  the  given  angle  B, 

3.  From  C  as  a  centre,  with  a  radius  equal  to  the  line  h 
describe  a  circle  cutting  BM  B.t  the  points  D  and  D\ 

Either  of  the  triangles  BCD  or  BCD'  will  then  be  the 
required  triangle. 

The  proof  follows  at  once  from  the  construction. 

189.  Scholium.  The  fact  that  there  may  be  two  tri- 
angles formed  from  the  given  data  has  been  explained  in  the 
echolium  §  113. 

Problem  XII. 

190.  Through  a  given  point  to  draw  a  straight 
line  which  shall  he  parallel  to  a  given  straight  line. 

Oiven.  A  straight  line,  AB)  2k  point,  P. 

Required.  To  draw  a  straight  line  through  P parallel  to  AB, 


PROBLEMS. 


88 


Construction.   1.  Take  any  point  D  in  AB,  and  join  PD 
2.  At  P  in  the  line  '^ 


p 


-B 


P2> .  make   angle   DPM  j^ 
equal  to  the  angle  PDB, 

3.  Produce  JE'P  in  the 
direction  F, 

EPF  will  then  be  the   .  _ 
required  straight  line  pass- 
ing through  P  parallel  to  AB. 

Proof.  By  Theorem  III.,  because  the  angles  PED  and 
PDB  are  alternate  angles. 

Problem  XIII. 

191.  To  divide  a  finite  straight  line  into  any 
given  numher  of  equal  parts. 

Given.  A  finite  straight 
line,  AB'y  a  number,  n. 

Required.  To  divide 
AB  into  n  equal  parts. 

Construction.  1.  IVom 
one  end  of  AB  draw  an 
indefinite  straight  line, 
making  an  angle  with  AB 
different  from  a  straight 
angle. 

2.  Upon  the  indefinite  line  lay  off  any  equal  lengths,  A  1, 
1  2,  2  3,  etc.,  until  n  lengths  are  laid  off. 

3.  Join  B  to  the  end  n  of  the  last  length. 

4.  Through  each  of  the  points  1,  2,  3,  ....  w,  draw  a 
parallel  to  nB  intersecting  AB. 

The  line  AB  will  then  be  divided  into  n  equal  parts  by 
the  points  of  intersection. 

Proof.  The  parallels  intercept  equal  lengths  on  the  trans- 
versal An,  by  construction. 

Therefore  they  also  intercept  equal  lengths  on  AB,  and 
the  number  of  intercepted  lengths  is  n  (§  132). 

Therefore  AB  is  divided  iuto  n  equal  parts.    Q.E.F. 


I 


84 


BOOK  11.    RECTILINEAL  FIGURES. 


Pkoblem  XIV. 

19J8.  T^o  adjacent  sides  of  a  parallelogra/nh  and 
the  angle  which  they  contain  being  given,  to  describe 
the  parallelogram. 

Given.  Two  lines,  ^Cand  Gil;  an  angle,  0, 

Required.  To  form  a  paral- 
lelogram having  GH  and  A  C  for 
two  adjacent  sides,  and  0  for  the 
angle  between  these  sides. 

Construction.  1.  At  one  end  ^ 'B 

A  of  the  line  A  C  make  an  angle  q. 
equal  to  0. 

2.  On  the  side  of  this  angle  take  AB  =  the  giren  line 

GH.  < 

3.  Through  B  draw  a  line  BD  parallel  Ui  AC. 

4.  Through  C  draw  CD  parallel  to  AB,  intersecting  BD 
in  D. 

ABCD  will  be  the  required  parallelogram. 
Proof.  May  be  supplied  by  the  student. 

193.  Corollary.  To  construct  a  square  upon  a  given 
straighi  line. 

This  problem  is  a  special  case  of  the  preceding  one.  in 
which  the  given  sides  are  equal  and  the  given  angle  is  a  xi^ht 

angle.     To  solve  it: 

DrswACrndBDlAB. 

Join  CD. 

And  the  square  will  be  complete. 


ind 
the 


DBMONBTRATION  OF  THEOREMS. 


85 


\ 

line 
BD 


wen 
.  in 


CHAPTER  VI. 

EXERCISES  IN  DEMONSTRATING  THEOREMS. 

The  following  theorems  should  be  demonstrated  by  the 
student  in  his  own  way,  so  far  as  he  is  able. 

Analysis  of  a  given  Theorem, 

The  first*  step  in  the  process  of  finding  a  demonstration 
is  to  state  the  hypothesis,  referring  to  a  diagram,  which  it  is 
generally  best  the  pupil  should  draw  for  himself.  The  state- 
ment should  include  not  simply  what  the  theorem  says,  but 
what  it  implies.  Beference  must  be  made  to  definitions  until 
the  hypothesis  is  resolved  into  its  first  elements. 

Next,  the  conclusion  must  be  analyzed  in  the  same  way, 
in  order  to  see  not  only  what  it  says,  but  also  what  it  implies. 

By  the  analyses  of  these  two  statements  they  must  as  it 
were  be  brought  together,  in  order  to  see  in  what  way  they 
are  related.  The  process  of  discovering  this  relation  is  one 
which  the  student  must  find  for  himself  in  each  case,  and  for 
which  no  rule  can  be  given.  "  Frequently,  however,  it  will  be 
necessary  to  draw  additional  lines  in  the  figure,  and  to  call  to 
mind  the  various  theorems  which  apply  to  the  figures  thus 
formed.  To  facilitate  this,  references  to  previous  theorems 
which  come  into  play  are  added. 

The  relation  being  found,  the  demonstration  must  next 
be  constructed  in  the  simplest  manner,  but  without  the  omis- 
sion of  any  logical  step.  This,  also,  is  a  matter  of  practice  in 
which  no  general  rule  can  be  given. 

It  is  recommended  that  the  teacher  require  the  pupil  to  express  each 
step  of  the  demonstration  with  entire  completeness  and  fullness.  Some 
of  the  first  theorems  are  so  simple  that  the  only  serious  exercise  is  that  of 
constructing  an  artistic  demonstration.  The  work  thus  becomes  a  valu- 
able exercise  in  language  and  expression  as  well  as  in  geometry. 

The  most  common  fault  is  that  of  passing  over  steps  in  the  demon- 
stration because  the  conclusion  seems  to  be  obvious.  One  of  the  great 
objects  of  practice  in  geometry  is  to  cultivate  the  habit  of  examining 
the  logical  fouadations  of  those  conclusions  which  are  accepted  without 


66 


BOOK  II.    RECTILINEAL  FIQURE8. 


critical  examination.  The  feeling  of  security  that  a  conclusion  is  right 
before  its  foundation  has  been  examined  is  a  most  fruitful  source  of 
erroneous  opinions;  and  the  person  who  neglects  the  habit  of  inquiring 
into  what  appears  obvious  is  liable  to  pass  over  things  which,  had  they 
been  carefully  examined,  would  have  changed  the  conclusion.' 

Remabk.  The  theorems  are  arranged  nearly  in  the  order  of  their 
supposed  difficulty.  The  references  give  the  theorems  on  which  the 
demonstration  n>ay  be  founded,  or  of  which  the  method  of  proof  has 
some  analogy.  It  is  not  to  be  expected  that  the  beginner  will  prove 
more  than  the  first  fifteen  or  twenty. 

Theorem  1.  If  a  line  be  divided  into  any  two  parts  and 
each  of  these  parts  be  bisected,  the  distance  of  the  points  of 
bisection  will  be  one  half  the  length  of  the  original  line. 

Theorem  2.  If  any  angle  be  divided  into  two  angles  and 
each  of  these  angles  be  bi- 
sected, the  angle  between  the  J) 
bisectors   will    be    half    the 
original  angle. 

Hypothesis.  BOD,  the  original 
angle  divided  into  the  two  angles 
JBOCand  COD  by  the  line  OC. 

OP,  OQ,  the  bisectors  of  BOO 
and  COD. 

Otmdusion.    Angle  POQ  =  i  angle  BOD. 

Theorem  3.  The  perpendicu- 
lars dropped  from  two  opposite 
angles  of  a  parallelogram  upon  the 
diagonal  joining  the  other  angles 
are  equal  (§§  128,  175). 

Theorem  4.  If  perpendiculars  be 
drawn  from  the  angles  at  the  base  of  an 
isosceles  triangle  to  the  opposite  sides, 
the  line  from  the  vertex  to  their  point 
of  crossing  bisects  both  the  angle  at  the 
vertex  and  the  angle  between  the  perpen- 
diculars. 

Theorem  5.  If  from  any  point  of  the  base  of  an  isosceles 
triangle  perpendiculars  be  dropped  upon  the  two  equal  sides, 
they  will  make  equal  angles  with  the  base. 


DEM0N8TBATI0N  OF  THEOREMS. 


87 


Thboeem  6.  If,  on  the  three  sides  of  an  equUateral  tri- 
angle,   points   equally  distant  from  the 
thi-ee  angles  be  taken  in  regular  order  and 
joined  bj  straight  lines,  these  lines  will 
form  another  equilateral  triangle  (§  108). 

Theobem  7.  If  the  perpendicular  from 
any  angle  of  a  triangle  upon  the  oppo- 
site side  bisects  this  side,  the  triangle  is 
isosceles. 

Theobem  8.  If  the  diagonals  of  any  quadrilateral  bisect 
each  other,  it  is  a  parallelogram  (§§67,  68,  108). 

Theorem  9.  If,  on  each 
pair  of  opposite  sides  of  a 
parallelogram,  we  take  two 
points  equally  distant  from  the 
opposite  angles  and  join  them 

by  straight  lines,  these  lines    ^  ^ 

^1   form   another   parallels  ^^.  ^^^«^ 

^^*  DT=.RB, 

v^rifr'"''^''  ^?*  ^  *^'  ^*'™"*«  ^"^'««  ^^^^^  by  a  trans, 
versa  crossmg  two  paraUels  be  bisected,  the  bisectors  will  be 
parallel  to  each  other  (§§  68,  71).  • 

Theorem  11.  If  either  bisector  of 
an  interior  angle  between  two  paral- 
lels be  continued  until  it  meets  the 
opposite  parallel,  it  forms  the  base 
of  an  isosceles  triangle  of  which  the 
equal  sides  are  the  transyersal  and 
the  intercepted  part  of  that  parallel. 

Corollary,  The  two  bisectors  of  the  angles  which  a  trans- 
versal makes  with  one  parallel  cut  ofE  equal  segments  of  the 
other  parallel  on  the  two  sides  of  the  transversal. 

Theorem  12.  If  the  four  interior  angles  formed  by  a  trans- 
v^al  crossing  two  paraUels  be  bisected,  and  the  bisectors 
produced  until  they  meet,  what  figure  will  be  formed  ?    (8  82^ 


his  theorem  is  to  be  enunciated  by  the  student 


wmmm 


88 


BOOK  II.    REOTILINEAL  FIOURES. 


Theoeem  13.  If  the  bisectors  of  the  four  interior  angles 
of  a  parallelogram  be  continued  until  each  one  meetstwo 
others,  they  will  form  a  rectangle. 

Theorem  14.  A  line  drawn  from 
any  point  of  the  bisector  of  an  angle 
parallel  to  one  side,  and  meeting  the 
other  side,  will  form  an  isosceles  tri- 
angle. 

Eypothem.  Angle  AOG  =  angle  COB. 

CP  II  BO. 
Conclusion.    PO  =  PC. 

Theorem  16.  If  any  two  interior 
angles  of  a  triangle  be  bisected  and 
a  line  parallel  to  the  included  side  be 
drawn  through  the  point  of  meeting 
of  the  bisectors,  the  length  of  this 
parallel  between   the   sides  will  be 
equal  to  the  sum  of  the  segments 
which  it  cuts  off  from  the  sides. 
Eypothem.    Angle  BAG  =  angle  BAB. 
Angle  BBC  =  angle  BBA. 
MBN  II  AB. 
Conclusion.    MIf=:  MA  +  NB. 

Theorem  16.  In  an  antiparallelogram— 

I.  The  angles  at  the  ends  of  the  upper  side  are  equal. 

II.  The  sum  of  each  pair  of  opposite  angles  is  equal  to  a 
straight  angle. 

III.  The  diagonals  are  equal  to  each  other  (§  172,  Rem.). 

Theorem  17.  That  portion  of  the  middle  parallel  of  a 
trapezoid  which  is  intercepted  between  the  diagonals  is  equal 
to  half  the  difference  of  the  parallel  sides  (§  170,  Del). 

Theorem  18.  If  the  diagonals  of  a  trapezoid  are  equal,  it 
is  an  antiparallelogram. 

Theorem  19.  The  sum  of  the  diagonals  of  any  quadri- 
lateral is  less  than  i^,he  sum  of  the  four  sides,  but  greater  thaix 
tlio  half  Sum. 


DEMONSTRATION  OF  THEOREMB. 


89 


Theorem  20.  If  from  any  point  in  the  base  of  an  isosceles 
tnangle  a  parallel  to  each  side  be  drawn  until  it  meets  the 
,  other  side,  the  sum  of  these  parallels  will  be  equal  to  either  of 
j  the  equal  sides  (§  121). 

Theorem  21.  If  the  middle  points  of  the  sides  of  any 
quadrilateral  be  joined  by  straight 
lines,  these  lines  will  form  a  paral- 
lelogram (§  136). 

If  the  given  quadrilateral  has 
its  pairs  of  adjacent  sides  equal  (cf. 
§  173),  the  parallelogram  formed 
from  it  will  be  a  rectangle;  and  if 
it  is  a  rectangle,  the  parallelogram  formed  from  it  will  be  a 
rhomboid. 

Theorem  22.  If  one  of  the  equal  sides  of  an  isosceles 
triangle  be  produced  beyond  the  vertex,  and  the  exterior  angle 
thus  formed  be  bisected,  the  bisector  will  be  parallel  to  the 
base  of  the  triangle. 

Theorem  23.  If  the  middle  points  of 
any  two  opposite  sides  of  a  quadrilateral  be 
joined  to  each  of  the  middle  points  of  the 
diagonals,  the  four  joining  lines  will  form  a 
parallelogram  (§  136). 

Theorem  24.    If  one   diagonal   of   a 
quadrilateral  bisects  both    of   the  angles 
between  which  it  is  drawn,  the  other  diagonal  will  cross  it  at 
right  angles. 

Theorem  25.  If  from  the  right  angle  of  a  right-angled 
triangle  a  perpendicular  be  dropped  upon  the  hypothenuse, 
the  two  triangles  thus  formed  will  be  equiangular  to  the 
original  one. 

Theorem  26.  If  one  of  the  acute  angles  of  a  right-angled 
triangle  be  double  the  other,  the  hypothenuse  will  be  double 
the  shortest  side. 


90 


BOOK  n.    BBOTILIITEAL  FIQUBE8. 


Theorem  27.  Each  side  of  any  triangle  is  less  than  half 
the  sum  of  the  throe  sides. 

Theorem  28.  If  one  side  of  an  isosceles 
triangle  be  produced  below  the  base  to  a  cer- 
tain length,  and  an  equal  length  be  cut  oft 
above  the  base  from  the  other  equal  side  and 
the  two  ends  be  joined  together  by  a  straight 
line,  this  line  will  be  bisected  by  the  base. 

Hypoth ma.    AG  =  BC;  AB  =  BF, 

Conclusion.    EN=  NF. 

Theorem  29.  The  sum  of  the  three 
straight  lines  drawn  from  any  point 
within  a  triangle  to  the  three  yertices  is 
less  than  the  sum  of  the  sides,  but 
greater  than  half  their  sum. 

Theorem  30.  If  from  the  yertex 
of  any  triangle  two  lines  be  drawn,  one 
of  which  bisects  the  angle  at  the  ver- 
tex, and  the  other  is  perpendicular  to 
the  base,  the  angle  between  these  lines 
will  be  half  the  difference  of  the  angles 
at  the  base  of  the  triangle. 

Theorem  31.  If  from  any  point  inside  of  an  equilateral 
triangle  perpendiculars  be  dropped  upon  the  three  sides,  their 
sum  will  be  equal  to  the  perpendicular  from  the  vertex  upon 
the  base. 

What  corollary  may  be  deduced  from  this  theorem? 

Theorem  32.  If  from  two 
opposite  angles  of  a  parallelo- 
gram lines  be  drawn  to  the 
middle  points  of  two  opposite 
sides,  these  lines  will  divide  the 
diagonal  joining  the  other  angles  into  three  equal  parts  (§137). 

Theorem  33.  If  from  either  angle  at  the  base  of  an 
isosceles  triangle  a  perpendicular  be  dropped  upon  the  opposite 
side,  the  angle  between  this  perpendicular  and  the  base  will  be 
one  half  the  angle  at  the  vertex  of  the  triangle. 


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,      BOOK  III. 
THE    CIRCLE. 


CHAPTER    1. 


GENERAL  PROPERTIES  OF  THE  CIRCLE. 


Major  conjugate  are. 


Definitions. 

1^4.  Bef,    The   oiroumference     Minor  conjugate  mo. 
of  a  circle  is  the  total  length  of  the 
curve-line  which  forms  it. 

195.  Def,  An  aro  of  a  circle  is 
a  part  of  the  curve  which  forms  it. 

196.  Def.  When  two  arcs  to- 
gether make  an  entire  circle,  they 
are  called  coi^iigate  arcs,  and  the 
one  is  said  to  be  the  coAJiigate  of  the  other. 

197.  Def.  When  two  conjugate  arcs  are  equal, 
each  of  them  is  called  a  semicirole. 

198.  Def.  When  two  conjugate  arcs  are  unequal, 
the  lesser  is  called  the  minor  arc,  and  the  greater  the 
major  aro. 

199.  Def    A  chord  is  a 

straight  line  between  two  points  — 
of  a  circle. 

200.  Def  A  secant  is  a 

straight  line  which  intersects  a 
circle. 


02 


BOOK  m,     THE  CinOLB. 


Remark.  A  secant  may  be  considered  as  a  chord  with  one 
or  both  of  its  ends  produced,  and  a  chord  as  that  part  of  a 
secant  contained  within  the  circle. 


circle  is  a  chord 


201.   Def,   The  diameter  of  a 

which  passes  through  its  centre. 

20%.  Def.  A  segment  of  a 
circle  is  composed  of  a  chord  and 
either  of  the  arcs  between  its  ex- 
tremities. 

303.  Def.  A  sector  is  formed 
of  two  radii  and  the  arc  included 
between  them. 

To  a  pair  of  radii  may  belong  either 
of  the  two  conjugate  arcs  into  which 
their  ends  divide  the  circle. 

304.  Def.  Concentric  olroles 

are  those  which  have   the   same 
centre. 

305.  Def     A  tangent  to  a 

circle  is  any  straight  line  which 
touches  the  circle  without  intersecting  it. 

306.  Special  Axioms  relating  to  the  Circle. 

I.  A  circle  has  only  one  centre. 
II.  Every  point  at  a  distance  from  the  centre  less 
than  the  radius  is  within  the  circle. 

III.  Every  point  at  a  distance  from  the  centre 
equal  to  the  radius  is  on  the  circle. 

IV.  Every  point  at  a  distance  from  the  centre 
greater  than  the  radius  is  without  the  circle. 

Theorem  I. 

307.  Circles  of  equal  radii  are  identically  equal. 
Hypothesis.   Two  circles  of  which  0  and  P  are  centres,  and. 

Eadius  OQ  =  radius  PR. 


GENERAL  PROPERTIES. 


Conclusion,   The  circles  are  identically  equal. 

Proof.  Apply  the  one  circle  to  the  other  in  such  manner 
that  the   centre  0  shall 
coincide  with  P,  and  OQ 
with  PR.     Then— 

1.  Because 0^=  Pi?, 
Point  Q  =  point  R. 

2.  Because  each  point 
of  the  one  circle  is  at  the 
distance  OQ  from  the  centre,  it  will  fall  on  the  other  circle. 

(§  206,  Ax.  III.) 
Therefore  the  circles  are  identically  equal.    Q.E.D. 


Theorem  II. 

208.  Equal  arcs  of  equal  circles  are  identically 
equals  subtend  equal  angles  at  centre^  and  contain 
equal  chords. 

Hypothesis.  AB,  CD,  

equal    arcs    around    the 

centres  0  and  P. 

0A  =  OB  =  PC=PD. 

Conclusion.  The  an- 
gles A  OB  and  CPD  and 
the  chords  AB  and  CD  are  equal. 

Proof.  Apply  the  sector  OAB  to  the  sector  PCD  so  that 
the  centre  0  shall  fall  on  P,  and  the  radius  OA  on  the  radius 
PC.    Then— 

1.  Because  OA  =  PC, 

Point  A  =  point  C. 

2.  Because  the  radii  are  all  equal,  every  part  of  the  arc  AB 
will  fall  on  some  part  of  the  circle  to  which  the  arc  CD 
belongs  (§  206,  III.). 

3.  Because  arc  AB  =  a.vc  CD,  the  point  B  will  fall  on  D, 
and  chord  AB  =  chord  CD.     Therefore 

Chord  AB  =  chord  CD. 


Angle  AOB  =  angle  CFD, 


Q.E.D. 


94 


BOOK  lU,     THE  OIROLE. 


Theorem  III. 

209,  Equal  angles  between  radii  irmlude  equal 
arcs  on  the  circle  and  eqvAil  cJiords, 

Hypothesis.    OA,  OB,  OP,  OQ, 
four  radii  of  a  circle  such  that 
Angle  AOB  —  angle  FOQ. 
Conclusions. 

Arc  AB  =  arc  PQ. 

Chord  AB  -  chord  PQ. 

Proof.  Apply  the  sector  A  OB  to 

the  sector  PO^  in  such  manner  that 

OA  shall  coincide  with  OP,  Then — 

1.  Because  OA  =  OP, 

Point  A  =  point  P. 

2.  Because  angle  A  OB  =  angle  POQ, 

OB  =  OQ. 

3.  Because  OB  =  OQ, 

Point  B  =  point  Q, 
Therefor3  AB  =  PQ.    Q.7il.D. 

4.  Because  all  the  radii  are  equal,  the  arcs  will  coincide 
between  P  and  Q. 

Therefore  the  arcs  are  identically  equal.    Q.E.D. 

210,  Corollary.  Sectors  of  equal  angles  %n  equal  circles 
are  identically  equal,  and  every  line  in  the  one  sector  is  iden- 
tically equal  to  the  corresponding  line  in  the  other  (§  174). 

Lemma: 

211,  A  sum  of  two  arcs  at  the  cmtre  subtends  an  angle 
equal  to  the  sum  of  the  angles  which 
each  arc  subtends  separately. 

Proof.  The  arc  ^P  subtends 
tH  f.agle  A  OB,  the  arc  BC  sub- 
tends the  angle  BOC,  and  the  arc 
ABC  subtends  the  angle  A  OC.  But 
A  DC  IB  by  definition  the  sum  of  the 
angles,  and  ABO  iu  the  sum  of  the  arcs* 
^omma. 


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GJSNERAL  PROPERTIES, 


d5 


The  Measurement  ol  Angles  by  means  of  Arcs. 

312.  From  the  preceding  theorems  it  follows  that  to 
every  arc  of  given  length  in  a  given  circle  corresponds  a 
definite  angle,  and  to  every  angle  corresponds  a  definite  length 
of  arc.  To  express  corresponding  arcs  and  angles  in  the 
shortest  way,  we  call  the  arc  corresponding  to  the  angle  ^0-B 
the  arc  angle  A  OB,  and  we  call  the  angle  corresponding  to 
an  arc  the  angle  arc. 

Thus,  in  the  following  figure. 

Angle  AOB=z  angle  arc  AB, 
Angle  A00  =  angle  arc  ABO, 
Arc  ABO=:  arc  angle  A  00, 
Atg  AB  =z  aro  angle  A  OB. 
Combining  Theorem  III.   with 
the  above  lemma,  it  follows   that 
arcs  can  be  taken  as  the  measure 
of  the   corresponding  angles,    and 
vice  versa. 

In  the  figure  the  circle  is  divided 
into  eight  sectors,  and  since  ^^°  = 
45°,  each  v^f  these  sectors  subtends 
an  angle  of  45°.    Therefore 

Angle  AOB=    45°;  arc  AB 

arc  ABO 
arc  A  BOD 
arc  ABODE 
arc  ABODEF 
arc  ABODEFG 


Angle  ^6>C=:  90° 
Angle  AOD  =z  135° 
Aj\glQAOE=  180° 
Angle  ^Oi^=  225° 
Angle  ^06'=  270° 
Angle  J  0^=  315° 
Angle  ^0-4  =  360° 


O 

=  46°. 
=  90°. 
=  135°. 
=  180°. 
=  226°. 


=  270°. 
arc  ABCDEFOH  =  315°. 
arc  ABODEFQHA  =  360°. 

913.  The  following  are  the  principles  to  which  we  are 
thus  led: 

I.  In  the  same  circle  or  in  equal  circles,  the  greater  arc 
measures  the  greater  angle. 

II.  A  minor  arc,  or  an  arc  less  than  a  semicircle,  measures 
an  angle  less  than  a  straight  angle. 


■\.:miji.m<jJMJuiiii i  ,mi 


96 


BOOK  III.     THE  OIRCLE. 


III.  A  major  arc  measures  a  reflex  angle  <yr  one  greater 
than  a  straight  angle, 

IV.  When  an  arc  measures  an  angle,  the  conjugate  arc 
tneasures  the  conjugate  angle, 

V.  The  sum  of  an  arc  and  its  conjugate  measures  the 
sum  of  an  angle  and  its  conjugate,  and  each  sum  is  a  circum- 
ference, or  360^. 

The  use  of  arcs  to  express  angles  has  a  great  advantage  of  making 
plain  to  the  eye  the  difference  between  an  angle  and  its  conjugate, 
because  we  can  always  draw  either  of  two  conjugate  arcs  between  the 
sides  of  the  angle. 

When  we  say  "  the  angle  AOF,"  we  should  not,  without  some  means 
of  distinction,  know  which  of  the  two  conjugate  angles  is  meant. 

But  when  we  say  the  angle  arc  ABCDBF,  we  do  know  which  of  the 
two  conjugate  angles  is  meant,  because  the  arc  measiu-es  only  one 
of  them,  not  both. 

When  we  do  not  use  arcs,  the  angle  expressed  without  any  adjective 
will  mean  the  lesser  of  the  conjugate  angles. 

When  we  mean  the  greater  conjugate  angle  and  do  not  use  an  arc, 
we  shall  call  it  a  reflex  angle. 


Theorem  IV. 

214.  In  a  circle  equal  chords  sv^teTid  equal  arcs 
and  equal  angles  at  the  centre. 

Hypothesis,    CD,  MN,  two  equal         ^ 
chords  of  a  circle  having  its  centre 
at  0. 

Conclusion, 

Angle  COD  =  angle  MON, 
Arc  CD  =  arc  MJST. 

Proof    In  the  triangles  COD  and 
MON  we  have 

CD  =  MN,  by  hypothesis. 

OC  =  ON,)  , 

OD  =  OM  \  "®^*^s®  *^®y  ^^®  '^^ii  of  the  circle. 

2.  Therefore  these  two  triangles  are  identically  equal. 

3.  Therefore 

Angle  COD,  opp.  CD  =  angle  MON,  opp.  equal  side  MN, 

4.  Therefore  arc  CD  =  arc  MN(§  209).     Q.E.D. 


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(§  115) 


GENERAL  PROPERTIES.  q„ 

i  ^ 

Theorem  V. 

major  arc, 

hypothesis,  AB,  CD,  two  chorda 
of  a  circle  such  that 

CD  >  AB, 
Conclusion. 
Minor  arc  CD  >  minor  arc  AB. 
Major  arc  CBAD<  major  arc  BCDA 

Proof.    From  the  centre  of  the  tT^ 

circle  draw  the  radii  OA    OB,  OC,  OD.    TheL 
1.  In  the  triangles  A  OB  and  COD  we  have 
0A  =  OC;  OB  =  OD'  AB  <:  m 
Therefore  angle  AOB  <  angle  COD.  ,. 

Therefore  minor  axe  AB  <  minor  arc  CD  (§  213, 1.). 

3.  Because  the  major  and  minni.  o^«  *«    j.i.         Q*"^*-^' 

Theore.aV.    Th'e^^mllt'bT^i  "*'*'"' ^"'"^^^ 

Theorem  VI. 

. ,  ^}j^'^^^ery  diameter  divides  the  circle  Mn  t..^ 
Identically  equal  semicircles.  ^  ^^^ 

hypothesis.    AMBN,  a  circle*  -^ ~"^ 

AB.a,  diameter;  0,  the  centre       ' 

Conclusion. 
Arc  AMB  =  arc  AI^B.  x\ 

Proof.    Draw  any  two  radii  OM  ^ 

^^i  ^^  making  equal  angles  with 
-^/3.  Turn  the  semicircle  AMB 
over  on  ^^  as  an  axis.    Then— 


98 


BOOK  III     THE  CIROLB. 


1.  Because  angle  ^OJf=  angle  ^OJV, 
^  Radius  0M=  radius  ok. 

2.  Because  OJf  =  ON, 

o    Q.         ..         Point  Jf=  point  JVT. 

3.  Since  M  may  be  any  point  whatever  on  the  circle 
every  point  of  the  arc  AMB  will  fall  on  a  point  of  ANB       ' 

Therefore  the  two  arcs  coincide  and  arridentically  equal. 

Q.E.D. 
Theorem  VII. 

cJtrp\frf  "^^^^^^^^  equally  distant  from  the 
centre^  and  of  unequal  chords 

the  greater  is  nearer  the  centre. 

Hypothesis.  AB,  CD,  and  MN, 
chords  of  a  circle  such  that 
AB  =CD>MN;  0,  the  centre 
of  the  circle;  OP,  OQ,  OR,  perpen- 
diculars from  0  on  AB,  CD,  and 
MN,  respectively. 

Conclusions.    I.  OP  =  OQ. 
II.  OP  <  OR. 

Proof.    I   Draw  the  radii  OA,  OB,  OC,  OD.    Then- 

1.  In  the  tnangles  ^  05  and  COD  we  have 
^dius  Oc=^  OB;  OD^  OA',  CD^AB  (hyp.). 

2.  Therefore  these  triangles  are  identically  equal,  and  the 

II.  Turn  the  figure  ORMN,  composed  of  the  chord  and  its 
peipendjou  »r,  around  the  centre  0  in  such  manner  that  oS 
'^r^fr'^t-^'  '''^  '^  the  point  in  Which  S^ 

3.  Because  the  radii  are  equal,  M~  A 

4    Because  MN  <  AB,  it  subtends  a  less  minor  arc  f8215^ 
and  the  point  J^T  will  fall  within  the  minor  arc  AB       ^^      ^' 

6.  Therefore  JfJVr  will  fall  inside  the  minor  segment  ^5 
aad  00  <  OR.  ^  ' 

6.  But  because  OP  is  a  perpendicular  on  AB, 

Ot^  <  00. 


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GENERAL  PB0PBBTIB8. 


99 


Comparing  6  and  6, 

OP  <  OR.     Q.E.D.  ' 

Theorem  VIII. 

Hypothesis.  CD,  a  chord  meet- 
ing a  circle  in  the  points  C  and  /), 
and  not  passing  through  the  centre. 

Conclusion.  CD  is  less  than  a  di- 
ameter. jDf 

Proof.  Let  0  be  the  centre  of 
the  circle.  Join  OC  and  OD,  and 
continue  />C?  across  the  centre  to  B, 
Then— 

1.  Because  CD  is  a  straight  line, 

CD<OC+OD. 

2.  Because  OCand  OB  are  radii, 

0C7+  00  =  BO-^OD  =  a,  diameter. 

3.  Comparing  (1)  and  (2), 

Diameter  >  CD.    Q.E.D. 

Theorem  IX. 

231.  T'^e  perpendicular  from  the  centre  of  a 
circle  upon  a  chord  bisects  the  chord  and  the  arc 
which  contains  it. 

Hypothesis.  CD,  a  chord  of  a  circle; 
Oy  its  centre;  OPQ,  a  perpendicular 
from  0  on  CD,  cutting  the  circle  in  v\ 

Conclusion.     PC=  PD. 

Arc  CQ  =  arc  QD. 

Proof.    Join  OC  and  Oi>.     Then— 

1.  Because  OC  and  OD  are  radii, 
they  are  equal. 

2.  Because  the  triangles  COP  and 
i>OP  have  0(7  =  OD,   OP  common. 


100 


BOOK  III.     THE  CIRCLE. 


and  OPG  and  OPD  right  angles,  they  are  identically  equal. 

Therefore  PO=PD.     Q.E.D. 

And  Angle  COP  -  angle  DOP. 

3.  Because  the  arcs  CQ  and  QD  are  subtended  by  the 
equal  angles  GOQ  and  DOQ,  ^ 

Arc  CQ  =  arc  QD,    Q.E.D. 

Theoeem  X, 

232.  Comersely,  a  line  bisecting  a  chord  at  right 
angles  passes  through  the  centre. 

Hypothesis.  CD,  a  chord  of  a  circle; 
PM,  its  perpendicular  bisector. 

Conclusion.  The  centre  of  the  circle 
lies  on  the  line  PM. 

Proof.  1.  Because  C  and  B  are  each 
upon  the  circle,  the  centre  is,  by  defini- 
tion, equally  distant  from  C  and  D. 

2.  Because  PM  is  the  perpendicular 
bisector  of  CD,  every  point  equally 
distant  from  C  and  D  lies  upon  this 
line  (§105). 

p  J*  TJ'S'^1^'''*®  *^®  ""^^"^^^  ""^  *^®  ^'''^^®  ^ies  upon  the  line 
JrM.     Q.E.D, 

Theorem  XL 

223.  Parallel  chords  or  secants  intercept  eaual 
arcs  between  them.  i'     ^^^ 

Hypothesis.  AB,  CD,  parallel 
straight  lines,  of  which  the  first 
meets  the  circle  in  the  points  A 
and  B,  and  the  second  in  the  points 
CsLiidD. 

Conclusion.  Arc  AC  =a.rc  DB. 

Proof.  Let  0  be  the  centre  of 
the  circle.  From  0  drop  the  radius 
OF  perpendicular  to  one  of  the  parallels.    Then— 


IM 


OBNEBAL  PB0PEMTIB8. 


101 


1.  Because  OF  is  perpendicular  to  one  of  the  paraUels,  it 
will  be  perpendicular  to  the  other  also  (§  72). 

2.  Because  OF  is  perpendicular  to  AB, 

AToAF=aTGBF. 

3.  Because  O^is  perpendicular  to  CD, 

Arc  CF=a,rcDF, 

4.  Subtracting  (2)  from  (3), 

AiG  AC  =  arc  DB,    Q.E.D. 


am) 


Theoeem  XII. 

234.  Of  lines  passing  through  the  md  of  any 
radius  the  perpendicular  is  a  tangent  to  the  circle 
and  every  other  line  is  a  secant  * 

Hypothesis.  0,  the  centre  of  a 
circle ;  OP,  a  radius ;  MN,  a  line 
through  P  perpendicular  to  OP; 
MS,  any  other  line  through  P. 

Conclusion.  I.  JfiV  is  a  tangent  | 
to  the  circle  at  P. 

II.  MS  is  a  secant. 

Proof  I.  1.  Because  OP  is  a 
perpendicular  from  0  upon  JfiV,  it 
is  less  than  any  other  line  from  0  to  JfJV  (§  101). 

2.  Therefore  every  other  point  of  MJVia  farther  from  the 
centre  0  than  P  is. 

3.  Therefore  every  point  of  ifiV  except  Pis  outside  the 
circle,  while  P  is  on  the  circle  (§206,  Ax.  III.). 

4.  Therefore  M]^  is  a  tangent  (§  205,  def .).      Q.  E.  D. 
Proof  II.    From  0  drop  a  perpendicular  upon  MS,  and 

let  Q  be  the  point  of  meeting.     Then— 

6.  Because  the  line  PM  is,  by  hypothesis,  different  from 
PM,  and  0PM  is  a  right  angle,  the  angle  0PM  cannot  be  a 
right  angle. 

6.  Therefore  OP  is  not  a  perpendicular  upon  MP. 

7.  Therefore  OQ,  the  perpendicular,  will  be  a  different 
line  from  OP. 


102 


BOOK  III     THE  CIRCLE, 


8.  Because  OQ  is  a  perpendicular,  it  will  be  less  than  OP 
an  oblique  line  (§  101).  ' 

9.  Therefore  the  point  Q  is  inside  the  circle  (§206,  Ax.  II. ). 
10.  Therefore  i^iS*  is  a  secant  of  the  circle.     Q.E.D. 

225.  Corollary  1.  The  radius  to  the  point  of  contact  of 
any  tangent  is  perpendicular  to  the  tangent. 

336.  Corollary  2.  The  perpendicular  from  the  point  of 
tangency  passes  through  the  centre  of  the  circle. 

Theorem  XIII. 

227.  Thco  tangents  drawn  to  a  circle  from  the 
same  external  point  are  equal,  and  make  equal  angles 
with  the  line  joining  that  point  to  the  centre. 

Hypothesis,  P,  a  point  out- 
side a  circle;  PM,  PN,  two 
tangents  from  P  touching  the 
circle  at  M  and  N-,  0,  the  cen- 
tre of  the  circle. 

Conclusion,    PM  =  PN, 
Angle  OPJf=  angle  OPN, 
Proof,     1.  In  the  triangles 
OMP  and  ONP  we  have 

Side  OJf=side  ON, 
Side  OP  =  side  OP, 
Angle  OMP  =  angle  ONP  =  right  angle. 
2.  Therefore  these  two  triangles  are  identically  equal; 
namely,  the  side  PM  is  equal  to  its  corresponding  side  PN, 
and  the  angle  0PM  to  OPN.    Q.E.D. 


OP, 

[I.). 

^  of 
tof 


the 
les 


al; 


INaOBIBBD  AND  CIRCUMSORIBED  FIGURES. 

CHAPTER     II. 

INSCRIBED  AND  CIRCUMSCRIBED  FIGUR'£S 


103 


Befinitions. 

^S^\''  ^^A  ^  rectilineal  figure  is  said  to  be  in- 
acnbed  in  a  circle  when  all  its  ver- 
tices lie  on  the  circle.    The  circle  is 
then  said  to  be  circumscribed  about 
the  figure. 

^  239.  Def.  A  figure  is  said  to  be 
circumscribed  about  a  circle  when        x  \       /  / 
aJU  Its  sides  are  tangents  to  the  circle.         ^<^^^^^^^ 

The  circle  is  then  said  to  be  in-  ^^Sf^'^^^i  polygon  and 

-,^_.^^  -  .      ,,      ^  ^^    ^    "^    "*        *  circumscribed  circle. 

scribed  in  the  figure. 

330.  Def.  An  inscribed  angle 

is  one  of  which  the  sides  are  two 
chords  going  out  from  the  same 
point  on  a  circle. 

231.  Def.  An  inscribed  angle  is  \x 

said  to  stand  upon  the  arc  included  a  circumscribed  polygon 
Detween  the  ends  of  its  sides.  *"**  *"  iMcribed  ciicie. 

If  the  sides  of  the  inscribed  angle  are 
PA  and  PC,  and  the  circle  is  divided 
into  two  segments  by  the  third  chord  A, 
A  G,  the  angle  APC'ib  said  to  be  inscribed 
m  the  segment  ACBPA  and  to  stand 
upon  the  arc  AMO, 

232.  Def.  A  line  is  said  to  sub-  ^-.„.^^ 

tend  a  certain  angle  from  a  certain  «« 

point  when  the  lines  dmwn  from^.'S\nd"2^  a^lS! 
tne  point  to  the  ends  of  the  line  ^  ^  *^«  segment 
form  that  angle.  t^^Auc.  '*~~  "^"^  *^^ 

The  line  ^C  subtends  the  angle  ^PCfrom  the  point  P. 


104 


BOOK  HI.     TUB  CIBOLB. 


Theorem  XIV. 

333.  If  from  any  point  on  a  circle  lines  he  drawn 
to  the  end  of  a  diameter  and  to  the  centre,  the  angle  at 
the  end  of  the  diameter  will  be  half  that  at  the  centre. 

Hypothesis.   AB,  &  diameter  of  a 
circle;  0,  the  centre;  P,  any  point  on 
the  circle. 
Conclusion. 

Angle  PBO  =  i  angle  POA. 
Proof.   1.  Because  OP  and  OB  are 
radii,  they  are  equal.     Therefore  the 
triangle  POB  is  isosceles;  whence 
Angle  OPB  =  angle  PBO. 

2.  POA  is  an  exterior  angle  of  the  triangle  POB.  There- 
fore 

Angle  OPB  +  angle  PBO  =  angle  POA.       (§  76) 

3.  Comparing  (1)  and  (2), 

Angle  P^O  =  i  angle  PO^.    Q.E.D. 

Theorem  XV. 

334.  JEach  angle  between  a  chord  and  the  tangent 
at  its  end  is  measured  by  half  the  arc  cut  off  by  the 
chord  on  the  corresponding  side. 

Hypothesis.  AB,&  tangent  touch- 
ing the  circle  at  T;  TO,  a  chord  from 
TtoC. 

Conclusions.  1.  Angle  A  TC  —  \ 
angle  of  arc  TA'C  ora  the  side  A. 

2.  Angle  BTC  =  ^  angle  of  arc 
TB'DC  on  the  Bide  B.  ^  T 

Proof.  Let  0  be  the  centre  of  the  circle.  From  T  draw 
the  diameter  TOD,  and  join  OC.  Suppose  the  chord  from 
Tto  fall  between  TO  and  TA.     Then— 

1.  Because  TA  is  a  tangent  and  TO  a  radius,  ATD  ia  a 
right  angle.    Therefore 

Angle  BTC—  right  angle  -f-  angle  OTO. 
Angle  ATC=  right  angle  —  angle  OTG. 


on 
at 
re. 


.B 


re- 


re) 


nt 
he 


)m 
I  a 


Or 


INSCRIBED  AND  CIR0UM8CRIBED  FIGUUEQ,       I05 

2.         Angle  TOC  =  straight  angle  TOD  -  angle  COD, 

=  2  right  angles  -  angle  COD. 
Reflex  angle  TOG=  straight  angle  TOD  -f-  angle  COD, 
g    g  =2  right  angles  +  angle  COD, 

Angle  COD  =  2  angle  OTC,  (8  233) 

4.  Comparing  (2)  and  (3), 

Angle  TOC  =  2  right  angles  -  2  angle  Ora 
Keflex  angle  TOC  =  2  right  angles  +  2  angle  C?ra 
6.  Comparing  with  (1), 

Angle  TOC  =  2  angle  A  TO. 
Reflex  angle  TOC  =2  angle  -5^(7. 

Angle  ATC=zi  angle  TOC, 

=  i  angle  arc  TM'C. 

Angle  BTC  =  i  reflex  angle  TOC, 

=  i  angle  arc  TB'C.     Q.E.D. 


Theoeem  XVI. 

335.  An  inscribed  angle  is  one  half  the  angle  of 
the  arc  on  which  it  stands. 

Hypothesis.  TC,  TD,  two 
chords  meeting  at  a  point  Ton 
a  circle. 

Conclusion.  Angle  CTD  = 
i  angle  of  arc  CD  (that  arc  be- 
ing taken  on  which  Tdoes  not 
lie).  ^~  r^ 

fi,  ^T^^i'^t.?^^*^^  ''®''*^®  ^^  *^®  circle.     From  0  draw 


1.  Angle  C7!5  =  |  angle  of  arc  CDB'T;  ) 

2.  Angle  Z>rj5  =  ^  angle  of  arc  DB'T. ' ) 


__„ _^.    ,  (§234) 

3.  Subtracting  the  second  equation  from  the  first,  and 
remarking  that 

Angle  CTB  -  angle  DTB  =  angle  CTD, 
Arc  CDB'T-  arc  i?5'r  =  arc  C2> 
we  have  angle  CTD  =  i  angle  of  arc  CD  whioh  rine^  «^t  '- 
dude  ^.    Q.E.D.  '  "*' 


106 


BOOK  III.    THE  OmOLB, 


Corollary  1.  The  angle  of  the  arc  CD  is,  by  definition,  the 
angle  COD  between  the  radii  OC  and  OD,  Therefore  the 
preceding  theorem  may  be  expressed  thus:  \ 

236.  7/*,  from  two  points  on  a  circle^  lines  be  drawn  to 
the  centre  and  to  any  third  point  on  the 
circumference,  the  angle  at  the  centre  will 
be  double  the  angle  at  the  circumference. 

But,  in  applying  the  theorem,  the  an- 
gle at  the  centre,  COD,  must  be  counted 
round  in  such  a  direction  as  not  to  include 
the  radius  07*  to  the  angle  at  the  cir- 
cumference. This  angle  will  therefore  be 
greater  than  180°  whenever  the  are  CTD  is  J'^S^"'^-^ 


a  mmor  arc. 


segment,  are  equal 


237.  Corollary'^,  All  angles  in- 
scribed in  the  same  segment  are  equal,  be- 
cause they  are  all  halves  of  the  same  angle 
at  the  centre, 

238.  Corollary  Z.  All  angles  inscribed 
in  a  semicircle  are  right  angles. 

For  they  are  all  measured  by  half  a  semi- 
circle. 

239.  Corollary  4.  If  a  triangle  be  in- 
scribed in  a  circle,  its  angles  will  divide 
the  circle  into  three  arcs. 

The  angle  of  each  of  these  arcs  will  be 
double  the  opposite  angle  of  the  triangle, 

240.  Corollary  5.  Every  pair  of  an- 
gles inscribed  in  conjugate  segments  are 
supplementary. 

For  if  ACB  and  AFB  dio  jwa  such 
angles,  each  of  them  is  mer ';iiirf''>  br  one  ^ 
half  the  opposite  arc,  and  th^j-cfore  their 
sum  is  half  a  circumference,  or  a  straight 
angle;  whence  they  are  supplementary,  by  definition  (§  60). 


'% 


I 


a 
a 


the 
the 


I  to 


T'D, 
ame 


'« 


B 


—-•»-«=•  ^ 


A"' 


INSORIBBD  And  CIRCUMSOnitiED  FIGURES.        107 

Theorem  XVII. 

4  ^^  J:  7!^^^^0h  three  given  points  not  in  the  sarne 
stratyht  line,  one  circle,  and  only  one,  may  be  dravm^ 

Hypothesis.  Ay  B,  C,  three 
given  points. 

Conclusion.  There  is  only  one 
point,  0,  so  siti'iiicd  that  it  may 
be  the  rontre  oi:  a  circle  passing 
through  tliean  points. 

Frnnf.  The  centre  of  the  circle 
must  be  equally  distant  from  A,  B, 
and  C. 

JoinAB  and  BC,  and  let  the  lines  m  and  n  be  the  per- 
pendicular  bisectors  of  AB  and  BC.    Then— 

1.  Every  point  which  is  equally  distant  from  A  and  B  lies 
on  the  Ime  m  (§  105). 

2.  Every  point  which  is  equally  distant  from  B  and  C  lies 
on  the  line  n. 

3.  Therefore  every  point  which  is  equally  distant  from 
all  three  points,  A,  B,  and  C7,  lies  on  both  the  lines  m  and  ni 
that  18,  on  their  point  of  intersection  0. 

4.  But  there  is  only  one  point  of  intersection.     Therefore 
there  is  one  point,  and  only  one,  equally  distant  from  A  B 
and  C;  namely,  the  point  0.  *     ' 

6.  Because  OA  =  OB  =  OC,  if  with  the  centre  0  and  the 
radius  OA  we  describe  a  circle,  it  will  pass  through  A   B 
and  a     Q.E.D.  e        ^     » 

iJvliolium.  If  the  three  points  A,  B,  and  C  are  in  a 
straight  line,  the  perpendiculars  to  the  lines  AB  and  BG 
are  parallel  (§  70).  Therefore  in  this  case  no  point  can  be 
found  which  shall  be  equally  distant  from  A,  B,  and  O. 

Theorem  XVIII. 

242.  m-om  any  point  within  a  circle  every  diam- 
eter subtends  an  angle  greater  than  a  right  angle, 
and  from  any  point  without  the  circle  it  subtends  an 
angle  less  than  a  right  angle. 


tiii 


108 


BOOK  III.     THE  CIRCLE. 


Hypothesis.     AB,  any  diameter  of  a  circle;  P,  any  point 
within  the  circle;  Q,  any  point  without 
the  circle. 

Conclusions. 
I.  Angle  APE  >  right  angle. 
II.  Angle  AQB  <  Hght  angle.        -^f 

Proof.  I.  Continue  either  side  of 
the  angle  APE,  say  AP,  until  it  meets 
the  circle.  Let  R  be  the  point  of  meet- 
ing. Join  BR.  ^^ 

Then— 

T.  Because  the  angle  APE  is  an  exterior  angle  of  the  tri- 
angle ERPj  it  is  greater  than  the  interior  angle  PRB  (§  77). 

2.  Because  the  angle  PRB  —  ARE  is  inscribed  in  a  semi- 
circle, it  is  a  right  angle  (§  238). 

3.  Therefore  APE  is  greater  than  a  right  angle.     Q.E.D. 
II.  The  proof  of  this  case  is  so  near  like  that  of  case  I. 

that  it  is  left  as  an  exercise  for  the  student. 

Theorem  XIX. 

243.  If  a  quadrilateral  he  inscribed  in  a  circle^ 
the  sum  of  each  pair  of  opposite 
angles  is  two  right  angles. 

Hypothesis.  AECD,  a  quadrilateral 
of  which  the  four  angles  lie  on  a  circle. 

Conclusions. 
Angle  A  -}-  opposite  angle  (7=2  right 

angles. 
Angle  E  +  opposite  angle  D  =  ^  right 
angles. 

Proof.  Draw  the  diagonal  ED.    This  diagonal  will  be  a 
chord  dividing  the  circle  into  two  segments.     Then — 

1.  Angle  BCD  =  i  angle  arc  BAD.  (§  235) 
Angle  EAD  =  l  angle  arc  BCD. 

2.  Adding  those  two  equations,  j 


=  2  right  angles. 


:>mt 


B 


tri- 

mi- 

.D. 
9 1. 


He, 


e  a 


35) 


i 

imCRIBEl)  AND  CmaUMaVIUBED  FIQUIiES.        109 

In  the  same  way,  by  drawing  the  diagonal  A  G,  may  be  shown 
Angle  B  -\-  angle  C=%  right  angles.     Q.  E.  D. 

Theorem  XX. 
244.  Comer sely,  if  the  sum  of  two  opposite  angles 
of  a  quadrilateral  is  equal  to  two  right  angles,  the 
four  angles  lie  on  a  circle.  — -^     ' 

Jiypothesis.    ABGD,  a  quadrilateral  D/1 9AC 

in  which 

Angle  A  -\-  angle  0=  two  right  angles.    ' 
Conclusion.   The  points  J,  ^,  C7and  a' 
B  lie  on  the  same  circle. 

Proof.    Describe  a  circle   through 
the  three  points  B,  A,  B.  ^ 

nr^nn''  T^^^^'f  ^^^* pass  through  C, it  must  intersect BC, 
or  i>C7  produced,  at  some  other  point  than  C. 
Let  Q  be  this  point.    Join  BQ.     Then— 

1.  Because  the  quadrilateraU5^Z>i8  inscribed  in  a  circle. 

Angle  BAB  -f-  angle  BQB  =  two  right  angles. 

2.  Angle  5^i>  +  angle  BOB  =  two  right  angles  (hyp.). 
Therefore  Angle  ^^Z)  =  angle  I  (7A 
Which  IS  impossible,  because  BQB  is  an  exterior  angle  of  the 
triangle  BQC  (§  77).  ^ 

3.  In  the  same  way  it  may  be  shown  that  the  circle  cannot 
intersect  BO  produced  at  any  point  beyond  0.  Therefore 
the  circle  must  pass  through  (7,  and  ABCB  lie  on  one  circle. 

Q.E.D. 
345.  Corollary.    Bach  exterior   angle  of  an  inscribed 
quadrilateral  is  equal  to  the  opposite  interior  angle. 

Theorem  XXI. 
S46.   When  two  chords  of  a  circle  intersect  each 
other,  each  angle  is  measured  by      a     — 
half  the  sum  of  the  arcs  interested 
by  its  sides  and  the  sides  of  its  ver-   /  X  p  ^d 

tically  opposite  angle. 

Hypothesis.     AB,    CB,   two  chords  Ca /n 

intersecting  at  the  point  F, 


110 


BOOK  III.     THE  CIRCLE. 


I    ■! 


Conclusions. 
Angle  DPB  z=  APC  =  ^  angle  arc  BD  -\-  ^  angle  arc  AO. 
Angle  APD  =  CPE  =  \  angle  arc  DA -\- ^  angle  arc  CB. 

Proof,    JomBC.    Then— 

1.  Because  APC  is  an  exterior  angle  of  the  triangle  BCP, 

Angle  APC  =  angle  PBC  +  angle  PCB.       (§  76) 

2.  Because  PBC  is  an  inscribed  angle  standing  on  the 
arc  A  C, 

Angle  PBC=  i  angle  arc  A C,  (§  235) 

3.  Because  PCB  is  an  inscribed  angle  standing  on  the 
arc  BDy 

Angle  PCB  =  ^  angle  arc  BD. 

4.  Comparing  (2)  and  (3)  with  (1), 

Angle  APC=^  angle  arc  AC  +  ^  angle  arc  BD.   Q.E.D. 

5.  In  the  same  way  may  be  shown 

Angle  APD  =  i  angle  arc  DA  +  ^  angle  arc  CB.    Q.E.D. 

Corollary.  Since  vertically  opposite  angles  are  equal,  we 
conclude — 

24:1i.  The  sum  of  each  pair  of  vertically  opposite  angles  is 
measured  by  the  sum  of  the  corresponding  intercepted  arcs  on 
any  circle  which  includes  the  vertex  of  the  angle, 

Theoeem  XXII. 

248.  If  two  secants  he  drawn  from  a  point  out- 
side a  circle^  the  angle  between  them  is  measured 
by  half  the  difference  of  the  intercepted  arcs. 

Hypothesis.    PAB,  PCD,  two  secants  emanating  from 

the  point  P  without  a  circle,  and  intersecting  the  latter  at 

the  respective  points  A,  B  and (7,  D. 

Conclusion. 

Angle  APC  —\  angle  arc  BD 

—  ^  angle  arc  CA. 

Proof.    Through  A  draw  a  par-  **! 

allel  to  PD,  intersecting  the  circle  ..^ 

in  the  points  A.  and  P. 

Then— 

1.  Because  BP  is  a  transversal  of  the  parallels  FA  and  DP, 
Angle  APC  =  corresponding  angle  BAF, 


c 

i 

n 
I 


B 

to 


The 


J-  Because  BA^  i.  ^  ,^^  ^^,^  ^^^^.^^  ^^  ^^ 
Angle  5^/-=  ^  angle  arc  Bi; 

3.  Because   fti  "nV^ln  ?  ^^  "  *  '"^'^  "«=  ^-O- 
parallels  ^^and  GO,  intercepted   between    the 

/•Co^parinrSr^alar"---  « -3, 

Angle  AFC=  l  ansle  aro  nn      i 

t  angle  arc  £D  ^  ^  angle  arc  CA,    Q.E.D. 

Theorem  XXIII 

"PP^^^e  sides  is  eoualLf^!     -^ 
of  the  other  pair  *^ '"^  ^- 

^.  ^,  i?,  >y.  *^®  P°^^*s  s 

-4^  =  A8, 
3-  B«t.eha;e,ai2i;^''+^«+^«  +  ^^. 

'  '        -fUt*  AU,  u. 


(§337) 


112 


BOOK  III.     THE  CIRCLE. 


!   I 


CHAPTER    III. 

PROPERTIES  OF  TWO  CIRCLES. 


250.  Def,  Two  circles  are  said  to  touch  each  other, 
or  to  be  tangent  to  each  other,  when  they  meet  in  a 
single  point  but  do  not  intersect. 

Theorem  XXIV. 

251.  Two  circles  cannot  intersect  in  more  than 

two  points. 

Hypothesis.  0,  P,  the  centres  of  two  circles ;  Jf,  N,  Q, 
three  of  their  points  of  intersection. 

Conclusion.     The  hypothesis  is  impossible. 

Proof.  If  0  and  P  were  the  centres  of  two  circles  pass- 
ing through  Mf  iV,  and  Q,  then  the  two  points  0  and  P 
would  be  equally  distant  from  all  three  points  M,  iV,  and  Q, 
which  is  impossible  (§  241). 

Axiom  V. 

252.  If  the  distance  between 
the  centres  of  two  circles  is  greater 
than  the  sum  of  their  radii,  they  will 
not  meet  each  other. 

Theorem  XXV. 

253.  i)^  the  distance  of  the  centres  of  two  circles  is 
equal  to  the  sum  of  their  radii,  they  will  he  tangent 
to  each  other. 

Hypothesis.  0  and  P,  the  centres 
of  two  circles,  the  sum  of  whose  radii 
is  equal  to  the  line  OP. 

Conclusion.  The  two  circles  have 
one  point  in  common,  and  no  more. 

Proof  1.  On  OP  take  a  point  M,  such  that  OM  shall 
be  equal  to  the  radius  of  the  circle  whose  centre  is  at  0.  The 
point  ilf  will  be  on  that  circle  (§  306,  Ax.  III.). 


PROPBRTIES  OF  TWO  CIRCLES. 


113 

tone;  M?Z  ^eiTol"'  *^^  '""  "'  '"^^  ™^".  ^e  d,V 
.  both  circles.  i"   "'  ^  will  be  at  the  same  time  on 

have  ~/+^/>1^7],^~S'^*  ^-.  -  shall 

5.  But  because  PJf  and  PiJ  are  radii  ;f  the  same  circle 

PH  =  PM.  ' 

6.  Taking  (5)  from  (4), 

„   _  OJi>  OM. 

7.  Therefore  the  point  R  lies  without  the  circle  around  O 

BelSr^efintist?*!  T.*  «  -^Po^-  h^ul"^^. 
are  tangent  ^rhZetJIIsT  St  ""'^""*"^'  '"^^ 

Theorem  XXVI. 

t^eir  raau,  tUey  .ill  SZt  IZZSt'""''  ""^ 

Hypothesis.    0,  P,  the  centres 
of  two  circles;  OA,  a  radius  of  the 
greater  one,  on  the  line  OP-  PB 
fine!^'"''  «^  *^e  lesser,  on  the  same 

OA-BP<:OP<OA^BP  ^      _/ 

S:?i"  J'r"''"  -t-«ect  in^^o  points. 

f^-;Tp.  ^h?co^r.itir  ^^^^^^^^^  P-s 

the  same  as  ^^  <^  ujl -\.  bp  is  therefore 

2    T«V  ^^"^BPKOA^BP, 

^.  J^akmg  away  the  common  part  PP, 

rn,       ,  OB<OA. 

-i-nereforfi  On  io  7«„„  ^i,„-.  ji        ,. 

point  5  on  thed^re  pt  wlv^  ^^  °*  *?'  '"'^'>  «'  ""d  t^e 
nil.  circle  1-  IS  withm  the  circle  0  (§  306,  Ax.  II.). 


114 


BOOK  III.     THE  ClIiCLK 


;i!  I 


Continue  OP  until  it  intersects  the  circle  P  in  Q.  Then — 

3.  Because  BP  and  PQ  are  radii  of  the  same  circle,  the 
condition  OA  -  BP  <  OP  gives 

OA  -BP^BP<  OP-^PQ, 
or,  which  is  the  same  thing, 

OQ  >  OA. 
Therefore  the  point  Q  is  without 
the  circle  0. 

4.  Because  the  point  B  is  within  the  circle  0  and  the 
point  Q  without  it,  if  we  move  a  point  along  the  circle  P 
from  B  to  Q,  this  point  must  cross  the  circle  0. 

5.  But  there  are  two  ways  in  which  we  can  go  from  B  to 
Q;  namely,  around  either  semicircle.  Therefore  there  must 
be  at  least  two  points  of  intersection  of  the  circles. 

6.  There  cannot  be  more  than  two  such  points,  because 
two  circles  would  ihen  pass  through  the  same  three  points, 
which  is  impossible  (§  341). 

Therefore  there  are  two.     Q.E.D. 

Theorem  XXVII. 

'  255,  If  the  distance  of  centres  of  two  circles  is 
equal  to  the  difference  of  their  radiiy  they  will  touch 
each  other  in  a  single  point. 

Hypothesis.  0,  P,  the  centres  of 
two  circles  such  that  the  line  OP  is 
equal  to  the  difference  of  their  radii. 

Conclusion.  These  circles  touch  in 
a  single  point,  and  no  more. 

Proof.    1.  Produce  the  line  OPy 
and  on  it  take  a  point  M,  such  that 
PM  shall  be  equal  to  the  radius  of  the  circle  P.     M  will  then 
be  on  that  circle. 

2.  Because  OP  is  equal  to  the  difference  of  the  radii,  the 
point  M  will  also  be  on  the  circle  0. 

3.  Therefore  the  point  M  will  be  common  to  both  circles. 
Now,  take  any  other  point  R  on  the  greater  circle,  and 

join  OR  and  PR.     Then— 

4.  Because  OR  is  a  straight  line, 

OP-^PR>  OR. 


PROPSRTIES  OF  TWO  GIMGLES. 


115 

6.  Because  OR  and  OM  &ve  radii  of  fTi«  oorv,     •    , 
OM=OF-^  PM,  ^^  ^^^^^®'  a^d 

fi    n  .      '     OR=.OP^PM. 

part  0p7""^  with  (4)  and  taking  away  the  common 

7.  Because  PJf  is  a  radius  of  the  circle  P  an^  pp  • 
greater  than  this  radius,  the  pointi.  falls  iuhecird^^^ 

8.  Therefore  eyery  Doint  nf  i\..    •    i  ^^z?^^'  ^^-  ^^'^ 
Without  the  Cele  TKllX^iLr^Lr  ^^ 

Axiom  VI.  ^•■^•^* 

the  distance  of  the  centrett^sfb: 'elthr^'  ^"^"^''^  "^ 
greater  than  the  sum  of  the  radii 
or  equal  to  the  sum  of  the  radii   ' 

"  of  ttetd^  '""  -"^  S™^*^^  '"-  ">«  ^iff-nce 
or  equal  to  the  difference, 
or  less  than  the  difference 

the^eltef '■'  ""'^'^'"'=^  *^  ^"""^"g  ^'oUaries  from 

tern^ll/^HnLlX-  ^nT'fi^'f  ""^  *°"'"'  each  other  ex- 
lies  wholly  ruSS»  i?  •  '^^  "'""  <^  ^^^>  ^^'^  °°«  "''«'« 
whoUy  iSe  le  tttr         '  "^  ''''  '"""'^  "^  <§«««)  "  " 


116 


BOOK  III     THE  CIRCLE. 


258.  Corollary  2.  When  ttvo  circles  touch  each  other,  the 
two  centres  and  the  point  of  contact  are  in  the  same  sk^aight 
line. 

259.  Corollary  3.  Two  circles  cannot  touch  each  other  in 
more  than  one  point,  unless  they  coincide  so  as  to  form  but 
one  circle. 

Axiom  VII. 

260.  When  two  circles  inter- 
sect, the  straight  line  which  joins  the 
two  points  of  intersection  is  a  chord 
of  each  circle. 


I 


'  Theorem  XXVIII. 

261.  When  two  circles  intersect  each  other ^  the 
straight  line  Joining  their  centres  bisects  their 
common  chord  at  right  angles. 

Hypothesis.  0,  P,  the  centres 
of  two  circles  which  intersect 
each  other;  M,  N,  their  points 
of  intersection. 

Conclusion,  The  line  OP  bi- 
sects the  line  ilfiV  at  right  angles. 

Proof.   Let  R  be  the  middle 
point  of  the  chord  MN.    Through  R  draw  a  perpendicular  to 
the  chord.    Then — 

1.  Because  MN  is  a  chord  of  the  circle  P,  its  perpen- 
dicular bisector  will  pass  through  the  centre  P  (§  222). 

2.  Because  MN  is  a  chord  of  the  circle  0,  its  perpen- 
dicular bisector  will  pass  through  the  centre  0. 

3.  Therefore  the  perpendicular  bisector  passes  through  the 
centres  of  both  circles. 

4.  But  there  can  be  only  one  straight  line  between  these 
centres.  Therefore  the  straight  line  OP  bisects  JfiV  perpen- 
dicularly in  the  point  R.    Q.E.D. 

2G2.  Corollary.  Conversely,  the  perpendicular  bisector  of 
a  common  chord  passes  through  the  centres  of  both  circles. 


PROBLEMS. 


117 


I 


CHAPTER     IV. 
PROBLEMS  REUTING  TO  THE  CIRCLE. 


Hereafter  we  shall  generally  present  with  ea^h  problem  an 
analysis;  that  is,  a  course  of  reasoning  by  which  the  solution 
may  be  arnved  at.  In  an  analysis  we  generally  begin  by  sup- 
posing  the  problem  solved,  and  reasoning  out  the  conditions 
which  must  thus  be  fulfilled. 

It  is  expected  that  the  analysis  will  generally  enable  the 
student  to  supply  the  proof  himself,  since  the  latter  will  com- 
prise the  same  reasoning  as  the  analysis,  but  generally  in  the 
reverse  order.  o  j         i^o 

Problem  I. 

263.  To  find  the  centre  of  a  gwen  circle 

Given.  A  circle,  A  BOD. 

Required.  To  find  its  centre.    --^^^  ~^ 

Analysts.   The  perpendicu- 
lar  bisector  of  every  chord  of 
the  circle  passes  through  the  ^f 
centre  (§232).      Therefore  if 
we  draw  two  such  chords  and 

bisect  each  of  them  at  right  

angles,  the  centre  will  lie  on  each  bisector;  that  is,  it  will  be 
their  point  of  intersection.     Hence  the  following 
^J'^^^iructio^  Draw  any  two  chords  of  the  circle,  as  AB 

Bisect  each  of  these  chords  at  right  angles  (8  179,  Cor  ) 
the  c'rdr'  ""  "  "'"'  *'^^  ''''''''''  "^^  blVe  centre  of 

chor^d?*'Thf  f '''!•    V'  ""^^  "'""''"^y  ^^*^^"y  *°  ^raw  the 
cnords.     The  construction  mav  be  found  as  follnwo.   i^.^^ 

thirnf  ^  ""^  f'  T^'  "'  ^  ''^*^^^  ^^^^  ^^y  radius"describ^ 
the  arc  of  a  circle.    From  any  other  point  B,  with  the  same 


^,    f 


/\ 


118 


BOOK  III.     THE  CIROLE, 


radius  doscribe  another  arc  intersecting  the  first  at  the  points 
P  and  Q.  ^  t 

The  straight  line  PQ,  produced  if  necessary,  will  pass 
through  the  centre  of  the  circle. 

In  the  same  way  another  line  passing  through  the  centre 
can  be  found,  and  the  centre  itself  will  then  be  their  point  of 
intersection. 

Corollary.  Since  we  need  not  use  any  definite  portion  of 
the  circle  in  this  construction,  we  may  in  the  same  way  find 
the  centre  when  only  an  arc  of  the  circle  is  given.  Then  from 
this  centre  we  may  describe  the  whole  circle.  Hence  this 
construction  also  enables  us  to  complete  a  circle  of  which  an 
arc  is  given. 

Problem  II. 

265.  From  a  given  point  without  a  circle  to  draw 
a  tangent  to  the  circle. 

Given.  A  circle,  TCT'\  a 
point  P  outside  of  it. 

Required.  To  draw  through 
P  a  tangent  to  the  circle. 

Analysis.  Any  tangent  to 
the  circle  is  at  right  angles  to 
the  radius  drawn  to  the  point 

of  tangency(§325).  Therefore  

the  centre  of  the  given  circle, 
the  point  P,  and 
the  point  of  tangency 
will  be  at  the  vertices  of  a  right-angled  triangle. 

But  a  right-angled  triangle  is  that  inscribed  in  a  semi- 
circle (§  238).  Therefore  the  point  of  tangency  will  be  on  the 
circle  of  which  OP  is  a  diameter. 

Construction.  From  the  centre  0  of  the  given  circle  draw 
the  line  OP,  and  bisect  it  at  the  point  C. 

From  C  as  a  centre,  with  the  radius  GO  =  GP  describe  a 
circle  OTPT,  intersecting  the  given  circle  in  Tand  T\ 

JoinPrandPJ". 

The  lines  PTand  PT'  will  each  be  a  tangent  to  the  given 
circle  and  will  pass  through  P  as  required. 


PROBLEMS. 


119 


Pkoblem  III. 

Given.   A  circle  and  a  point  P  ^ 

upon  it. 

ii?e^i«>e^.    Through  P  to  draw  ai 
tangent  to  the  circle.  ( 

^w«/j^5w.   The  tangent  will  be  at 
right  angles  to  the  radius  from  the    V  7^1 

centre  to  P  (§  224).    Hence  we  have     ^-^^ 

tTlgh  p'  *'^^  ^^^"^  -^  ^--  «  perpendicular  to  it 

tiJrdTatiroS    rtr^^'  r  ^-^^-  ^- 

Join  AB.  '  """^  ""^  *^^  ^q"^^  ai'cs  i'^,  PP. 

Bisect  AB  at  right  angles  by  a  line  00 
fore^aShr^r^  '"^  Pe^endicular  to  OQ  and  there- 

^^S^^^^^  i^  =  fhLS  t^  ^S  ortht"  i'  'I 
bisects  the  arc  AB  (§§  221,  222)  ®  ^"^ 

^     ^-  .because  the  arcs  Pj' and  PP  are  eoual   Pic  fi,    v.-      . 
mg  point  of  the  arc  AB.    TlhevetaJnn'    1  *^^  ^''^^*- 

3.  Because  PTi.rZ     -^^f^^foi'e  0^  passes  through  P. 
gent  required  P^^P^^^^^-^^r  to  this  line,  it  is  the  tan- 

Scholium.    The  radina  nn  ,•«  ^  x 

conseruction,since^"i^for;Vr7S'°  *'^  """^ 

Problem  IV 

oil.  \  t^llTc^  -  «  ^--  ^-•-.^. 

Required.  To  inscribe  a  cir- 
cle within  it. 

Analysis.  The  bisectors  of 
the  ^three  angles  ^,  P,  and  (7 
mee.  in  a  point  equally  distant 
from  the  three  sides  of  the  tri-  A-  ^ 


1 

i 

m 

a  r  Mil 

Hlf 

J'l 

t. 

1 

, 

"  j 

ii 

i 

120 


BOOK  III.     TllK  CIIWLR. 


angle  (§  164).  Thoroforo  this  point  is  the  centre  of  the 
required  circle. 

Construction.  1.  Bisect  any  two  angles  of  the  triangle  as 
A  and  D  by  the  lines  AO  and  Z?0,  and  let  0  bo  their  point 
of  meeting. 

3.  From  0  drop  a  perpendicular  OD  upon  any  side,  as  AB. 

3.  From  0  as  a  centre,  with  the  radius  OD  describe  a 
circle.     This  circle  will  be  the  required  inscribed  circle. 

Proof.  The  perpendiculars  from  0  upon  each  of  the  three 
sides  of  the  circle  are  equal  (§  1G4). 

Therefore  each  of  these  sides  is  a  tangent  to  the  circle. 


if 


268.  Scholium.  In  the  general  triangle  (§  58)  there  are 
four  circles,  each  fulfilling  the  condition  of  touching  the  three 
sides  of  the  triangle.  One  of  these  is  within  the  triangle  as 
just  described,  and  the  other  three  are  without  it,  each  of 
them  touching  one  of  the  sides  from  without.  The  latter  are 
called  escribed  circles. 


rrprt!  xr 'iicrr  °' ''- ''''  '^'^ 

J^T^LI^S'  •-'  '"'  *•«  ^'^  -'-  ot  the  .^cHbod 
^  0'  bisector  of  exterior  angle  SAB 
^^^'        "        "        "         "      ^i9P. 

BCR. 
QOA. 


7iO" 
CO" 
CO"' 
AO'" 


n 

it 


(( 
(( 
tt 
ti 
t( 


Problem  V. 

269.  To    describe   a   circle   which     ^hnii    ^ 
through  three  given  points,  ^"^^  ^""'^ 

Given.     Three  points,  A,  B,  C. 

Required.  To  describe  a  circle 
which  shall  pass  through  each  of 
them. 

Analysis.     The  centre  of  the  .. 
circle  must  be  equally  distant  from 

This  circle  will  be  thaf.  r^f„,i.«^    ^.__-.       ., 
points  ^,  B,  and  a  '-a«J— ,   i^uBbing  througii  the 

Proof.     As  in  §§  166,  241. 


122 


BOOK  III     THE  CIRCLE. 


370.  Corollary.  Since,  in  the  construction  of  this  prob- 
lem, we  describe  a  triangle  having  its  angles  at  the  given 
points,  this  problem  is  the  same  as  that  of  circumscribing  a 
circle  about  a  given  triangle. 

Problem  VI. 

271.  To  bisect  a  given  arc  of  a  circle,  . 

Given.  An  arc  AB.  p 

Required.   To  bisect  it. 

Analysis.  1.  The  radius  which 
is  perpendicular  to  the  chord  of 
the  arc  bisects  both  the  chord 
and  the  arc. 

2.  The  line  which  bisects  the  ~  Q 
chord  at  right  angles  passes  through  the  centre  of  the  circle, 
so  that  that  part  of  it  which  is  contained  between  this  centre 
and  the  circle  itself  is  a  radius. 

3.  Therefore  this  line  will  bisect  the  arc  of  the  chord. 
Construction.  1.  Draw  the  chord  AB  between  the  two 

ends  of  the  given  arc. 

2.  Bisect  this  chord  at  right  angles  by  the  line  PQ. 

3.  Let  D  be  the  point  in  which  this  bisector  intersects 
the  given  arc  AB. 

The  point  D  will  bisect  the  arc. 


Problem  VII. 

272.   Upon  a  gimn  line  as  chord,  to  describe  an 
arc  of  a  circle  of  wMcTi  the  in- 
scribed angle  shall  be  equal  to 
a  given  angle. 

Given.  A  line,  ^-B;  an  angle,  X,  ^^ 

Required.  On  ^5  as  a  chc  rd,  to 
draw  an  arc  of  a  circle  such  that 
any  angle  inscribed  in  it  shall  be 
equal  to  X. 

Analysi'^.  1.  Suppose  the  whole 
circle  of  {     ch  an  arc  is  required  to 


PJiOIiLEMlS. 

123 

oho'nr^uppi:  *r ^"'  ^^^'  "^  "-^^  «'  0-  end  of  the 

dicular  bisector  of  the  chord  AB  will  both  „.T,  t^     ^"T"" 

centre  of  the  circle  (§§  223,  226).     Hence     ^  "^"^  **>' 

Construction.    1.   At  one  extremity  as  J   of  rt«  r       .  „ 

make  an  angle  BAD  ennni  tn  ti><,    •  .  '       '"e  ''"«  ^^ 

a.  At  yl  erect  a  perpendicular  ^0  to  the  line  ia 
The  ::tt  oVn^^  1  "^■'*  '-S'-  b/the  Le  W. 

^..e^^^oSrr^i;  7:r^'  -^-'-^ «' -^^^  «„ 

eide'Vith  Tb,  X:  "li^fi  r'^,  r  '■-  ^  ^^  -»  eoin. 
circle,  and  Oik  Ztvt  '''"  '''  "  ^'^'^''  "f  the 

Pboblem  VIII 

<y*vm.    A  circle;   a  ^ 

triangle,  ^^a 

Required.  To  in- 
scribe in  the  circle  a  tri- 
angle  which  shall  have 
Its  angles  equal  to  the 
respective  angles  A,  B, 
^,  of  the  triangle  ABC, 

Analysis.      1.    jSupl 
pose    the    triangle    in- 
scribed, and  let  it  be  ^ ' ^'r^'      Ai-  j^  a 
the  circle.  "^  ^^  c  .    At  A'  draw  a  tangent  BG  to 

^-  We  shall  then  hn^a 


\''\[ 


\:\  I 


i;!!!!'      ! 


H 

! 

^^^B^^B 

■  i  1   ■■ 

^^■H 

1  i; 

i' 

Hi 

124 


BOOK  IT!.     THE  CIRGLB. 


3.  Therefore  the  three  angles  of  the  triangle  will  be  the 
same  as  the  three  angles  DA'O'y  C'A'B',  B'A'C.    Hence 

Construction.  1.  Draw  a  tangent  to  the  circle  at  any 
point  A'. 

3.  At  the  point  of  tangency  A  make  the  angle  DA'C 
equal  to  the  angle  B  of  the  given  triangle;  the  angle  G'A'Bl 
equal  to  A'y  when  the  angle  B'A'G  will  be  equal  to  G  (§  73). 

3.  Produce  the  sides  until  they  intersect  the  circle  in  B' 
and  C",  and  join  B'G' 

A'B'G'  will  be  the  required  inscribed  triangle. 

Peoblem  IX. 

274.  Ahout  a  given  circle  to  circumscribe  a  tri- 
angle which  shall  he  equiangular  to  a  given  triangle. 

Given.  A  tri- 
angle, ABG;  a 
circle,  PQE. 

Required.  To 
circumscribe  about 
PQR  a  triangle 
which  shall  have 
angles  equal  to  A, 
Bf  Gj  respectively. 

Analysis. 

1.  Suppose  the  problem  solved.  Let  P,  Q,  R,  be  the  points 
in  which  the  sides  of  the  required  triangle  touch  the  circle, 
and  let  0  be  the  centre  of  the  circle. 

2.  Join  OP,  OQ,  OR. 

3.  In  the  quadrilateral  OPC'Q  we  have 

Angle  OqC'  =  right  angle; )  ;«  „^.. 

Angle  OPG'  =  right  angle.  S  ^^  '^'^^^ 

And  because  the  sum  of  all  four  angles  is  equal  to  four  right 
angles,  if  we  take  away  these  two  angles  from  all  four  angles 
we  have  left 

Angle  QOP  +  angle  QG'P  ~  two  right  angles. 

4.  Therefore  QOP  is  the  supplement  of  (7',  or  of  its  equal, 
G;  and  in  the  same  way  POR  is  the  supplement  of  B,  and 
ROQ  of  A. 


PROBLEMS. 


125 


4^ro^  ^z^  iiKr rf c/'"^ 

equ^  to  the  respective  angles  around  the  c&  0     Henor 

Construetvon.    1.  Produce  each  side  of  the  riven  tri»2> 
so  as  to  form  its  three  exterior  angles  ^  ^^^ 

3.  At  the  end  of  each  radius  draw  i  tnnffnT,*  +«  +1, 
a.d  produce  these  tangents  untilX  S^ril'^'/  ^^^^^^' 
This  triangle  will  be  that  required.  ^ 

Peoblem  X. 
ciS:  ^"^  ^^"""^  ""  ^"^"^^t^ngmt  to  two  given 

Given.     Two   circles     P^^^T^v — -—-^        o 

i'^  and   Q8,  of  which    "^    ^    ^  ~ ^ 

QS  is  the  lesser. 

Required,  To  draw  a 
straight  line  which  shall 
be  tangent  to  both  circles.  ^ 

8.  Let  ^  and  0  be  the  centres  of  the  circles. 

c  paidiieis  /^cy  and  PQ,  they  are  equal  (8  126^ 

^p':^X  ^i^:r'  *-  '•'^  ^'«— '"'^H;  radu 


'Mn 

f 
1 

J 

'    I 

fl 

126 


BOOK  III      THE  GUWLE. 


i    I 


il 


.pnf *  Jf>.     ^       '  T*''  ^  ^^  *^^  «^^"^^  ^ir^^^e  draw  a  tan- 
gent to  the  inner  circle,  and  let  B  be  the  point  of  tangencr 

3.  Join  AB,  and  produce  the  joining  line  to  P 

4.  From  the  centre  0  draw  the  line  OQ  parallel  to  AP 
and  meeting  the  circle  in  ^.  v  i^    c*  i^i  to  ^^ 

6.  Join  Pg.     The  line  PQ  will  then  be  one  of  the  tan- 
gents required.  ^" 

Proo/.     To  be  supplied  by  the  student  from  the  analysis. 

EXERCISES. 

Theorem  1.  If  two  chords  be  drawn  in  a  circle  intersect- 
ing each  other  at  right  angles,  their  ends  will  divide  the  circle 
into  four  pros.  Show  that  the  sum  of  each  pair  of  opposite 
ares  is  a  semicircle.  x-  ri-  »  ^^ 

How  will  the  theorem  be  modified  if  the  chords  do  not 
intersect  within  the  circle? 

Theorem  2.  If,  from  the  two  ends  of  a  chord,  chords  per- 
pendicular to  It  be  drawn,  they  will  be  equal  in  length. 

Theorem  3.  The  shortest  line  between  two  concentric 
circles  IS  part  of  the  radius  of  the  outer  one. 

Theorem  4.  If  the  angles  at  the  base 
of  a  circumscribed  trapezoid  are  equal, 
each  non-parallel  side  is  equal  to  half  the 
sum  of  the  parallel  sides  (§  227). 

Theorem  5.  If  from  the  centre  of  a 
circle  a  perpendicular  be  dropped  upon  either  side  of  an 
inscribed  triangle,  and  a  radius  be  drawn  to  one  end  of  this 
side,  the  angle  between  the  radius  and  perpendicular  will  be 
equal  to  the  opposite  angle  of  the  triangle. 

Theorem  6.  If  two  equnl  chords  intersect,  the  segments 
of  the  one  are  respectively  equal  to  the  segments  of  the 
otner. 

Theorem  7.  The  only  parallelogram  which  can  be  in- 
scribed  in  a  circle  is  a  rectangle. 

Theorem  8.  From  any  point  outside  a  circle  a  chord  sub- 
tends an  angle  less  than  half  its  arc;  from  any  point  inside 
the  circle  an  angle  greater  than  half  its  arc. 


THEOREMS  FOR  EXERCISE 


127 


which 


which 


^  Explain  the  relation  of  the  two  c( 
the  chord  divides  the  circle  to  the  sid 
the  subtended  angle  lies. 

Theoeem  9.    If  an  angle  between 
a  diagonal  and  one  side  of  a  quadri-  C 
lateral  IS  equal  to  the  angle  between 
the  other  diagonal  and  the  opposite 
side,  the  same  will  be  true  of  the 

three  other   pairs  of    angles  corre^' ^ 

spondmg  to  the  same  description,  and   ^^P'    -^OB^adb. 
the  four  vertices  of  the  quadrilateral    ^''^'  %ba-'c?jPa 
he  on  a  circle  (§237).  DAbZb%. 

TTTTTrtPTj-xr  in     A     •    ,     ,  ^^CZ)  on  a  Circle. 

Theorem  12.  If  a  circle  0  pass  " 
through  the  centre  of  another  circle 
Py  and  from  the  centre  of  P  a  di-  ^ 
ameter  to  the  circle  0  be  drawn 
every  chord  of  P  passing  (when 
produced)  through  the  other  end  Q 
of  this  diameter  is  bisected  by  the 
circle  0  (§§  221,  238). 

Theorem  13.  If  two  circles  be 
drawn  each  touching  a  pair  of  parallel 
mes  and  a  transversal  crossing  them, 
the  distance  between  the  centres  of 
the  circles  is  equal  to  the  length  of 
the  transversal  intercepted  between 
the  parallels  (§  227.) 

Theorem  14.  If  any  number  of  tri- 
ang  es  have  the  same  base  and  equal 

angles  at  the  vertices,  the  bisectors  of 
thes_e  angles  pass  through  a  noinf.  (^  9si<y\ 

iJ-ow  is  this  point  defined?"  ■'"''"'^* 


Condtision.    WN'=:  Ify, 


128 


BOOK  III     THE  GIRGLE. 


Theorem  15  Of  all  chords  passing  through  a  given  point 
within  a  circle,  the  least  is  that  which  is  bisected  by  the  point. 
I  Theorem  16.  The  centres  of  the 
four  circles  circumscribed  about  the 
four  triangles  formed  by  the  sides  and 
diagonals  of  a  quadrilateral  lie  on  the 
vertices  of  a  parallelogram  (§  166). 

Define  the  lengths  of  the  sides  of 
this  parallelogram  and  its  angles.  Af 

The  four  circles  circumscribed  about  the  triangles ^05,  BOG,  GOD 
DO  A,  have  their  centres  on  the  vertices  of  a  parallelogram. 

Theorem  17.  The  tangents  at  the  four  vertices  of  an  in- 
scribed rectangle  form  a  rhombus. 

Theorem  18.  The  quadrilateral 
formed  by  the  bisectors  of  the  four 
angles  of  another  quadrilateral  has 
its  four  vertices  on  a  circle. 

Theorem  19.  If  each  pair  of 
opposite  sides  of  an  inscribed  quad- 
rilateral be  produced  until  they 
meet,  the  bisectors  of  the  angles 
formed  at  the  point  of  meeting  will 
be  perpendicular  to  each  other. 

Theorem  20.  If  the  arc  cut  off 
by  the  base  of  an  inscribed  triangle 
be  bisected,  and  from  the  point  of 
bisection  be  drawn  a  radius  and  a 
line  to  the  opposite  vertex,  the 
angle  between  these  lines  will* be 
half  the  difference  of  the  angles  at 
the  base  of  the  triangle. 

Theorem  31.  If  perpendiculars  be  dropped  from  the  ends 
of  a  diameter  upon  any  secant,  their  feet  will  be  equallv  distant 
n-om  the  points  in  which  the  secant  intersects  the  circle 


129 


THEOREMS  FOB  EXERCISE. 

Theorem  33.  The  middle  points 
of  all  chords  passing  through  a  fixed 
point  lie  on  a  circle  (§§  136,  338). 


Theorem  33.  If  a  chord  be  ex- 
tended by  a  length  equal  to  the  radius, 
and  from  the  end  a  secant  be  drawn  « 
througli  the  centre  of  the  circle,  the 
greater  included  arc  will  be  three 
times  the  lesser. 

Hyp.  PB  =  radius. 

Cone.   AxcAN=^qxqBM. 


Theorem  34.  If  a  chord  be  pro- 
duced equally  each  way,  and  from  its 
ends  tangents  be  drawn  to  the  circle 
on  opposite  sides,  the  line  joining 
the  point  of  t?ngency  will  bisect  the 
chord. 


^.J^f^^''^^''  ^\  .?  a  right-angled  triangle  the  sum  of  the 
hypothenuse  and  the  diameter  of  the  inscribed  circle  is  equal 
to  the  sum  of  the  two  sides.  ^ 

Theorem  36  If  lines  be  drawn  from  the  centre  of  a  circle 
to  the  vertices  of  any  circumscribed  quadrilateral,  each  pair 
of  opposite  angles  at  ^  he  centre  will  be  supplementary. 

nirlTT.^^'  ^^  r  ^^^^l^^^^al  triangle  be  inscribed  in  a 
circle  the  distence  of  any  point  on  the  circle  from  the  farther 
side  of  the  triangle  is  equal  to  the  sum  of  its  distances  from 
the  two  nearer  sides. 

Theorem  38.  If  in  any  triangle 
the  feet  of  the  perpendiculars  from 
the  angles  upon  the  opposite  sides  be 

joined,  the  thi^e  angles  of  the  new . :s 

triangle  thus  formed  wiU  be  bisected  by  the  perpendiculars. 


r 
S 


BOOK  IV, 
OF  AREAS. 


Definitions. 

376.  Base  md  Altitude.  Def.  The  base  of  a  fig- 
ure is  tliat  one  of  its  sides  on  which  we  conceive  it  to 
rest. 

Any  side  of  a  figure  may  be  taken  as  its  base. 

377.  Def.  The  altitude  of  a  figure  is  the  perpen- 
dicular distance  of  its  highest  point  above  its  base. 

The  altitude  of  a  paral- 
lelogram is  the  length  of  the 
perpendicular  dropped  from 
any  point  of  one  side  to  the 

opposite    side,   produced    if  . 

necessary.     The  latter  side  is  then  considered  as  the  base. 

The  altitude  of  a  triangle  is  the 
length  of  the  perpendicular  dropped 
from  either  angle  to  the  opposite 
side,  produced  if  necessary  The 
latter  side  is  then  considered  as  the 
base. 

The  terms  hase  and  altitude  are  therefoi^T^^ativr  alti- 
tude meaning  a  perpendicular  distance  from  the  base. 

278.  Def.   The  perimeter  of  a  iiolvo-nn  \.  +i. 
combined  length  of  aU  its  sides  -"^o--  -  «« 


vxiss 


b 


AUlia  OF  RECTANGLES.  ^^^ 

279.  Def     A  rectangle  contained  by  two  lines 

IS  a  rectangle  of  which  two  adja-  g 

cent  sides  are  equal  to  these  lines. 
Tho  rectangle  contained  by  tlio  lines 
a  and  h  is  that  in  which  one  j)air  of 
opposite  sides  are  each  equal  to  a,  and 
the  other  pair  to  b. 

380.  Def.  The  projection  of  a 
finite  line  upon  an  indefinite  line  is 
the  distance  between  the  per- 
pendiculars dropped  from  the 
ends  of  the  finite  line  upon  the 
indefinite  line. 

Example.  ^'^' is  the  projec- 
tion of  the  line  AB  upon  the  in- 
definite fine  X, 

381.  Def.    The  area  of  a  ,,__    _ 
plane  figure  is  the  extent  of  A  ~Y 
surface  of  which  it  forms  the  boundary 

A  plane  figure  cannot  have  an  area  unless  it  completelv 
incloses  a  portion  of  the  plane  in  which  it  lies.         ''^"'P^^^^y 
Example  1.  An  angle  has  no  area. 
Ex.  2.  Two  parallel  lines  do  not  inclose  an  area. 
Jix.  3.  But  polygons  and  circles  have  areas. 

283.  In  elementary  geometry  we  regard  the  area  is  +hA 
^2-^1^  "  '"""  "*"  *""  °^  ""-«  P-'«  —  to 


132 


BOOK  IV.     OF  AliEAS. 


CHAPTER    I. 

AREAS  OF  RECTANGLES. 


Square  Inch. 


Square 
centimetre. 


S83.  Areas  are  measured  by  supposing  them  divided 
up  into  units. 

To  form  a  unit  of  area,  we  take 
any  unit  of  length  and  erect  a 
square  upon  it.  The  area  of  this 
square  is  the  corresponding  unit 
of  area. 

We  may  thus  form  a  square 
millimetre,  a  square  centimetre, 
a  square  inch,  etc. 

Lemma. 

284.  The  number  of  units  of  area  in  a  rectangle 
is  equal  to  the  'product  of  the  numhers  of  units  of 
length  in  its  containing  sides. 

Proof  for  whole  numbers.  In  the  figure  let  the  base  be  m 
units  in  length,  and  the  vertical 
sides  each  n  units.  Then  if  we 
divide  the  sides  into  units  of 
length  and  join  all  the  correspond- 
ing division  points  by  straight 
lines,  the  whole  rectangle  will  be  divided  up  into  units  of 
area.  It  is  evident  that  there  will  '»e  in  the  whole  rectangle 
n  rows,  each  containing  m  units.  Therefore  the  whole  num- 
ber of  units  wiU  be  mn. 

285.  Corollary.  The  area  of  a  square  of  which  each  side 
contains  m  units  is  m^  units. 

Note.  The  above  demonstration  presupposes  that  each  side  of  the 
rectangle  contains  a  whole  number  of  units.  The  general  case  in  which 
the  sides  contain  fractions  of  a  unit  must  be  deferred  until  after  the 
subject  of  proportion  is  taught. 


*!i 


aheas  of  rectangles. 


133 


1 


or 


286.  Scholium.  The  preceding  result  is  sometimes  ex- 
pressed by  saying  that  the  area  of  a  rectangle  is  equal  to  the 
product  of  two  of  its  containing  sides.  But  when  we  speak  of 
the  product  of  two  lines,  what  we  really  mean  is  the  product 
of  the  number  of  units  which  the  lines  contain.  This  product 
IS  always  equal  to  the  number  of  square  units  in  the  rectan- 
gular area.  Hence  we  may  consider  such  an  area  as  a  kind  of 
product  of  lines,  and  for  shortness  use  a  form  of  algebraic  pro- 
duct to  represent  it,  as  follows.     Instead  of  saying 

The  rectangle  contained  hy  the  lines  AB  and  AD, 
we  may  say 

Rectangle  AB  .  AD, 

Area  AB  .  AD, 

or  simply  AB  .  AD. 

If  the  lines  are  represented  by  single  letters,  as  a,  b,  wo 
may  write  simply  ab 

to  express  the  area  of  the  rectangle. 

To  these  expressions  of  algebraic  products  we  may  assign 
either  a  geometric  or  an  algebraic  signification. 

Geometrically,  the  expression 

AB .  CD 

memsthe  area  of  the  rectangle  contained  by  the  lines  AB 
and  CD,  This  meaning  may  be  considered  independently  of 
any  idea  of  a  product.  ^ 

Algebraically,  the  same  expression  means  the  product  of 
the  number  of  units  in  CD  by  the  number  of  units  in  AB. 

Since  this  product  is  equal  to  the  number  of  units  of  area 
m  the  rectangle,  the  two  meanings  are  entirely  consistent. 

Again,  geometrically,  the  expression 

AB^ 
means  j?Ae  area  of  the  square  erected  upon  the  line  AB 
^rrJ    I  Tz?""'"^  ""^y  ^®  ^''^'^^^  Without  any  idea  of  the 

geometncal  application,  we  should  write  or  understand  the  word  "area  » 
before  the  STmbola  of  nrodnnfo  ♦,.  „v^^  *v„x  .,  _        ,     ^"^^     ^^^ 

totiiinkof  themgeometricalIy7  ^  ""  "'  ""''^''''  ""  '^P'"*"^ 


10 


184 


BOOK  IV.     OF  ABEAS, 


I     :i. 


^i!  in  : 


i      ^l 


Theorem  I. 

28?.  T7ie  area  of  the  rectangle  formed  hy  tioo 
lines,  one  of  which  is  divided  into  several  parts,  is 
equal  to  the  sum  of  the  areas  of  all  the  rectangles 
formed  hytJieundivid'               (f       jf            r' 
ed  line  and  the  several  ^\  ^         ' ' ^ 


D 


4L_£. 


-% 


parts  of  the  divided 
line, 

Ilyiwthesis.      A    lino,   ®  ^0" 

BM=A;  another  line,  BF  =  P,  divided  into  the  parts  a, 
0,  Cy  d,  etc.,  ut  the  points  (7,  />,  E^  etc. 

Conclusion.  Area ^.P  =  area  (.!.<% +  ^.^)  +  j.^  +  yl.,n. 

Proof  On  BP  erect  the  rectangle  BPNM,  of  which  the 
sides  BM,  PN,  shall  each  be  equal  to  the  lino  A,  and  MJV 

""ni  YJ'^r.^:     ^^  ^'  ^^  ^'  ^^^-  «^^«^  *he  perpendiculars 
GO  ,  DD'y  EE'y  etc.,  meeting  MNm  C",  />',  W,  etc.   Then— 

1.  Because  the  angles  at  C,  D,  E,  etc.,  are  all  right 
angles,  each  of  the  quadrilaterals  MBCC,  C'CDD\  etc.  is  a 
rectangle.  ' 

2.  The  sum  of  all  these  rectangles  is  equal  to  the  rectande 
BM.  BP,  by  construction. 

3.  Therefore  area  BM.  BPz=  sum  of  areas  A.a^A  .1, 
etc.,  or,  because  BM  =  A,  and  BP  =  P, 

Area  A.P  =  area  {A.a  +  A.h  +  A*c  +  ^.^).     Q.E.D. 

Schohum.  When  we  give  the  symbols  A,  a,  h,  c,  d,  which 
we  have  supposed  to  represent  lines,  their  algebraic  significa- 
tion, we  have 

P  =  a-^l  +  c^d, 
which  gives  the  well-known  formula 

^{fi^-h  +  c-\-d)=:Aa-{-Al-\-AG-{-Ad, 
and  expresses  the  distributive  law  in  multiplication. 

Theorem  II. 

288.  If  a  straight  line  he  made  up  of  two  parts, 
the  square  of  the  whole  line  is  equal  to  the  sum  of  the 

squares  of  the  two  parts  plus  twice  the  rectangle  con- 
tained mt  flip  /M/y/r/o  ^ 


n 


AREAH  OF  mCTAmiEti.  jg^ 

ITypothesis.  AB,  a  straight  lino  dividnd  nf   p  •  *    .. 
parta  A  J*  and  PB,  aiviaed  at  P  into  the 

Condmion.    Square  on  AB 
oquuls  sum  of  squares  on  AP  G 
and  Pi^  plus  twice  the  rectan- 
gle ^P./'/?.     Or,  in  symbols, 

^B'=AP'-\.PB'-\.2AP,PB.  W 

Proof.    On  AB  erect  the 
square  A  BCD. 

On  AC  take  ^/"  =  ^7:>. 

Through  P  draw  PP" 
parallel  to  ^C7,  and  through  P' 
draw  P'P'"  parallel  to  AB, 
Then—  A^  ^ 

1.  Taking  away  from  the  equallines^;?  AP  ^\.^  ..      i 
parts  ^P,  AP\  we  have  '  ^  ^'  *^®  ^'^''^^ 

^^  =  ^'^. 
CQ.  DQ,  and  ^^  are'pi^^rl^f '  *''  quadnlaterak  ^ft 

ngM  angle.    Therefore  they  are  all  rectangles  (8  125) 

4.  Because  of  the  parallelism  of  the  lines  JcPP"  .„^ 
BB,  and  also  of  the  lines  ^BP'J'^^  l^lcuXZ^     ' 

P"'D  =   QP"  =  ptQ 
6.  Therefore  Aiea  APQP'  =  j^p\ 
Area  QP"'DP"  =  pp» 
Area  P'QP^'C  =  AP  '.  pp. 
Area  PBP"'Q  =  AP    pp 

aJ:  Thtrf °"'  """^  ""^«  "P  ^''o'  -«»  of  the  sqnare  on 
uence:  —   ~^  '^^-'  •  "^■'^» 


(1) 


'  ^  -f^v 


136 


BOOK  IV.     OF  AREAS. 


289*  If  frwu  the  square  of  the  sum  of  two  tinea  we  take 
away  the  sum  of  their  squares,  we  shall  have  left  twice  their 
rectangle. 

290.  Scholium.   By  hypothesis  we  have 

AB  =  AP-i-  PB. 
Substituting  this  in  the  conclusion,  wo  have 
(AP  +  PBY  =  AP^  +  'HAP  .  PB  +  PB\ 
a  well-known  algebraic  expression.  ^ 

The  geometric  construction  serves  to  exhibit  to  the  eye 
the  different  parts  of  which  this  algebraic  expression  is 
made  up. 


iii 


ill 


B 


K 


B- 


n 


Theorem  III. 

291.  T7i&  square  upon  the  difference  of  two  lines 
is  equal  to  the  sum  of  the  squares  upon  the  lines^ 
diminished  by  twice  the  rectangle  contained  by  them. 

Hypothesis.  AB,  AG,  two  lines  of  which  AC  ia  the 
longer;  BC,  their  difference. 

Conclust07i. 
BC  =  AB'  +  AG'  -  2AB .  AG. 

Proof.  On  AG  erect  the  square 
AGOH. 

On  ^C  erect  the  square  BGEF,  D 

On  AB  erect  the  square  ABKL. 

Produce  FE  till  it  meets  AOm 
D.     Then—  A 

1.  The  whole  area  AKLBGHG 
=  AB'  +  AG\ 

2.  Because  EB  —  BG,  and  BL 
:=  AB,yfQ  have 

EL  =  AB  -{-  BG  =  AG. 
Therefore 

Area  KLDE  =  area  AB  .  AG. 

3.  Because  CH  =  AG,  and  GF  =  BG,  we  have 

FH  =  AG -==  BG  =z  AB. 
Hence 

Area  DFQH  =  area  -4-5 .  -4  0. 


i«  we  take 
vice  their 


0  the  eye 
'ession  is 


jOo  lines 
\e  lineSf 
*y  them. 

C  is  the 
iH 


AliBA^H  OF  liEOTANQLEa. 


187 


4.  If  from  the  whole  area  (1)  we  take  away  the  areas  '2) 
and  (3),   we  have  loft  the  square  BVEF;   that  is.   BO* 
Therefore  >   -w^^  . 

292,  Scholium.  Since  BO  =z  AO  ^  AB,  vro  have 
(AO-^  ABy  =  A0'-  2AB.A0+AB\ 
the  algebraic  formula  for  the  square  of  the  difference  of  two 
numbers. 

TlIEOKEM    IV. 

293.  T7ie  difference  of  the  squares  described  on 
two  lines  IS  equal  to  the  rectangle  contained  by  the 
sum  and  difference  of  the  lines. 

Hypothesis.  -4  i?,  vlC,  two  straight  lines  of  which  ulC  is 
tiio  greater,  and  each  of  which  is  to 
have  a  square  described  upon  it.      % ~ ' iD 

Conclusion. 

AC'-~AB^=:(AC-{.AB){AO-AB).  p 

I'roof.  On  AO  describe  the 
square  A  ODE.  On  AE  take  AF= 
AB.  From  F  draw  FH  parallel  to 
AO,  meeting  OD  in  H,  and  from  B 
draw  ^(7 parallel  to  AE  and  meet- 
ing FH  in  Q.     Then— 

iV^^' vl^\^Tr''^  ^'  '^  *^'  '^'^  *^«°^^^  i*  °iay"be  shown 
that  EH  and  6^(7  are  rectangles,  and  ^  6^  a  square. 

2    Because  AE=  AO  {h^  construction),  and  AF  =  AB, 
to /a  Thirllf  ~  ^^^  ^"^  ^^  ^«'  ^^  --^-^-^  equal 
Rectangle  EH  =  AO  {AO -  AB). 

3.  Because  /'^  and  ^C  are  parallel,  OH  =  AF  =  AB 
While  5C ,s,  by  construction,  equal  to  AO^  AB.    Therefore 

Rectangl  GO  =  AB  (AO-  AB) 

4.  The  sum  of  the  rectangles  AO(AO~^AB)  and  ^5 


\  /  A  n 


yi/j). 


(^C  —  AB\  =  reotana-lo (  AH ^  a  p 

^r  i^  ^n^  ^^ffe^f  f«  between  the  squares  on  the  lines  i^  and 
AC  IS  made  up  of  the  sum  of  these  rectangles. 


-   I 


U 


m 


IH  < 


II 


138 


BOOK  IV.     OF  ABEAa. 


Therefore 

AC -- AB"  =  (AG -{•  AB)  (AG -- AB),    Q.E.D. 

294.  Scholium.   Expressing  the  areas  of  the  squares  and 
rectangles  in  algebraic  language,  this  theorem  gives 

a»  -  5»  =  (a  +  b)  (a  -  b). 


■•  ♦  > 


CHAPTER    II. 

AREAS   OF    PUNE  FIGURES. 


Theorem  V. 


395.  The  area  of  a  parallelogram  is  equal  to  that 
of  the  rectangle  contained  by  its  base  and  its  altitude. 


Hypothesis.  ABGD,  any  parallelogram  of  which  the  side 
ABiB  taken  as  the  base;  AE,  the  altitude  of  the  parallelo- 
gram. 

Gonclusion.   Area  A  BCD  —  rectangle  AB  .  AE. 

Proof.  From  A  and  B  draw  perpendiculars  to  the  base 
AB,  meeting  CD  produced  in  E  and  F.     Then — 

1.  Because  A  BCD  and  ABEF&re  both  parallelograms, 

EF  =  AB,  and  CD  =  AB.  (§  127) 

Therefore  EF  =  CD. 

BF  =  AE. 
BD  =  AG. 

2.  If  from  the  line  ED  we  take  away  EF,  FD  remains; 
and  if  we  take  away  CD,  EC  remains.  Because  the  parts 
taken  away  are  equal  (1), 

FD  =  EC. 


ARBA8  OF  PLANE  MOUBBS. 


139 


3.  Comparing  with  the  last  two  equations  of  (1)  it  is 
seen  that  the  triangles  BFD  and  AEC  have  the  three  sides  of 
the  one  equal  to  the  three  sides  of  the  other. 

Therefore 

Triangle  AEC  =  triangle  BFD, 
4   Prom  the  trapezoid  ABED  take  away  the  triande 
AEC,  and  there  is  left  the  parallelogram  ABCB,     From  the 
same  trapezoid  take  away  the  equal  triangle  BED  and  there 
18  left  the  rectangle  ABEE,    Because  the  triangles  are  equal 
Rectangle  ABEE  =  parallelogram  ABCD,  * 

Therefore 

Area  ABCD  =  rectangle  AB  .  AE.    Q.E.D. 

296.  Corollary  1.  All  parallelograms  upon  the  same  base 
and  between  the  same  parallels  are  equal  in  area,  because  thev 
are  all  equal  to  the  same  rectangle. 

297.  Cor.  2.  Parallelograms  having  equal  bases  and 
equal  altitudes  are  equal  in  area. 

4T.  ??^'  7  ^^^'  ^'     ^^  ^^^  parallelograms  having  equal  bases, 
that  has  the  greater  area  which  has  the  greater  altitude 

Of  parallelograms  having  equal  altitudes,  that  has  the 
greater  area  which  has  the  greater  base. 

Theorem  VI. 

399.  ^Ae  area  of  a  triangle  is  equal  to  half  the 
area  of  the  parallelogram  formed  from  any  two  of  its 
sides,  having  an  angle  equal  4o  that  betwem  those 


Hypothesis.    ABC,  any  triangle;  PQRS,  a  parallelogram 


im 


m  which 


J  if 
11 


140 


BOOK  IV.     OF  AREAS. 


PQ^  R8  =  AB, 
PR=  QS=AG. 
Angle  PQ8  =  angle  CAB. 
Conclusion.  Area  ABC  =  ^  area  PQR8, 

Proof.   Draw  the  diagonal  P8.     Then— 

1.  Because  of  the  equations  supposed  in  the  hypothesis, 
the  triangles  PQ8  and  ABC  have  two  sides  and  the  included 
angle  of  the  one  efqual  to  two  sides  and  the  included  angle  of 
he  other.     Therefore  the  triangles  are  identically  equal,  and 

Area  ABC  =  area  PQ8.  (§  108) 

2.  In  the  same  way  is  shown 

Area  ABC  =  area  PR8. 

3.  The  sum  of  the  areas  PQ8  and  PR8  makes  up  the 
whole  area  of  the  parallelogram  PQR8.  Therefore,  com- 
paring (1)  and  (2), 

Area  ABC^^  area  PQR8.    Q.E.D. 

300.  Corollary.  A  diagonal  of  a  parallelogram  divides 
it  into  two  triangles  of  equal  area. 

Theorem  VTI. 

301.  The  area  of  a  triangle  is  one  half  the  area 
of  the  rectangle  contained  by  its  base  and  its  altitude. 

Hypothesis.  ABC,  a  triangle  having  the  base  AB  and 
the  altitude  CD. 

Conclusion.  Area 
ABC^^VQctAB.CD. 

Proof.  Through  B 
draw  BG  parallel  to 
AC,  and  through  G 
draw  CG  parallel  to 
AB,  meeting  BGinG. 
Then- 

1.  ABCG  is  a  parallelogram  having  the  base  AB  and  the 
altitude  CD.    Therefore 

Area  ABCG  =  rect.  AB  .  CD.  (§  295) 

2.  Because  AB  and  CD  are  each  sides  of  this  parallelo= 
gram. 

Area  ABC  =  ^  area  ABGG.  (§  299) 


I 


AREAS  OF  PLANE  FIG  USES. 


141 


3.  Comparing  (1)  and  (2), 

Area  ABC  =  i  rect.  AB  .  CD.    Q.E.D. 

302.  Corollary  1.  All  triangles  on  the  same  base,  having 
their  vertices  in  the  same  straight  line  parallel  to  the  base,  are 
equal  to  each  other  in  area. 

303.  Cor.  2.  If  several  triangles  have 
their  vertices  in  the  same  point,  and  their 
bases  equal  segments  of  the  same  straight 
line,  they  are  equal  in  area. 

304.  Cor.  3.     If  a  triangle  and  a         —^ ^ 

parallelogram  stand  upon  the  same  base  and  between  the  same 
parallels,  the  area  of  the  parallelogram  will  be  double  that  of 
the  triangle. 

Theorem  VIII. 

305.  The  area  of  a  trapezoid  is  equal  to  that  of 
the    rectangle    con-  o 
tained  by  its  altitude 
and  half  the  sum  of 
its  parallel  sides. 

Hypothesis.  ABCD, 
a  trapezoid  of  which  the 
sides  AB  and  CD  are  -*  1 

parallel;  CE,  the  altitude  of  the  trapezoid. 

Conclusion.    Area  ABCD  =  ^{AB  +  CD) .  CE. 

Proof.    Draw  either  diagonal  of  the  trapezoid,  say  BC. 

Then — 

1.  Because  ABG  is  a  triangle  having  AB  as  its  base  and 
(7^  as  its  altitude. 

Area  ABC  =  ^AB  .  CE.  (§  301) 

3.  Because  BCD  is  &  triangle  having  CD  as  a  base  and  an 
altitude  equal  to  the  distance  of  the  vertex  B  from  CZ>— that 
is  (because  AB  and  CD  are  parallel),  to  CE— 
Area  BCD  =  i  CD  .  CE. 
3.  The  sum  of  these  areas  makes  up  the  whole  area  of  the 
i'i»i;cauiQ.     inereiore 

Area  of  trapezoid  =  ^AB .  CE  -f  ^CD .  CE 

=  UAB  +  CD)  CE{%  187).   Q.E.D. 


II'  il 
ii 


142  BOOK  IV.     OF  ARBAH 

Theoeem  IX. 

306.  If  through  any  point  on  the  diagonal  of  a 
parallelogram  two  lines  be  drawn  parallel  to  the 
sides,  the  two  parallelograms  on  each  side  of  the 
diagonal  will  be  equal. 

Hypothesis.  ABGD,  a  paraUelogram;  P,  any  point  on  the 
diagonal  AD  ;  RS, 
MN,  lines  passing 
through  P,  parallel 
to  AB  and  AG  re- 
spectively, and  meet- 
ing the  four  sides  in 
the  points  P,/^,  J/,  JV.  A' 


H 

T 


y-...^..,._^r....A 


X 


N 


Conclusion.    Area  RPMO  =  area  NB8P. 
Proof.    1.  Because  the  lines  AD,  AP,  and  PD  are  the 
diagonals  of  the  respective  parallelograms  ABGD  ANRP 
and  PSMDy  we  have  ' 

Area  ACD  —  area  ABD. 
Area  ARP  -  area  J iVP; )  ,„ 

Area  PMD  =  area  PSD.  S  ^»  ^^"^ 

3.  From  the  area  A  GD  take  away  the  areas  ARP  and 
PMD,  and  we  have  left  the  area  RPMG. 

3.  From  the  equal  area  ABD  take  away  the  equal  areas 
ANP  and  PSD,  and  we  have  left  the  area  NB8P.  There- 
fore 

Area  RPMG  =  area  NBSP.    Q.E.D. 

307.  Definition.  In  the  foregoing  constructions  the 
parallelograms  ANRP  and  PSMD  are  called  paraUelo- 
grams  about  the  diagonal  AD. 

RPGM  and  NBP8  are  called  the  complements  of 
parallelograms  about  the  diagonal  AD, 

Theoeem  X. 

308.  In  a  right-angled  triangle  the  sauare  of 
the  hypothenuse  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides. 


AREAS  OF  PLANE  FIGUBEB. 


143 


Hypothesis.  ABO,  a  triangle,  right-angled  at  A:  BAGF, 
AGKH,  EC  ED,  squares  on 
its  respective  sides. 

Conclusion, 
Area  BA  GF  -f  area  A  CKH  F 
=  area  BCED, 

Proof,  Through  A  draw 
AL  parallel  to  BD  and  CE, 
meeting  DE  in  L.  ,  Join  FG 
and  AD. 

The  proof  will  now  be 
an'anged  as  follows: 

We  shall  show  (1)  that 
the  triangles  FBG  and  ABD 
are  identically  equal;  (2)  that 
the  area  BAGF  is  double  that  of  the  triangle  FBG;  (3)  that 
the  area  BL  is  double  that  of  the  equal  triangle  ^5Z).  From 
this  will  follow  area  ABGF=  area  BL.  It  may  be  shown  in 
the  same  way  that  the  area  of  the  square  on  AG  is  equal 
to  that  of  the  rectangle  GL,  from  which  the  theorem  will 
follow. 

1.  In  the  triangles  ABD  and  FBG  we  have 

BA  =  BF,),     .       , ,     . 
BD  =  BO  \  ^  Wothesis. 

Angle  DBA  =  right  angle  DBG+  angle  ABO, 
Angle  FBG  =  right  angle  FBA  +  angle  ABO. 
Therefore  the  two  triangles  haying  two  sides  and  the  in- 
cluded angle  of  the  one  equal  to  two  sides  and  the  included 
angle  of  the  other  are  identically  equal,  so  that 

Area  ABD  =  area  FBG. 

2.  Because  BA  G  and  BAG  are  both  right  angles  (hypoth- 
esis), GA  and  A  0  are  in  the  same  straight  line. 

Therefore  the  triangle  FBG  is  on  the  same  base  FB,  and 
between  the  same  parallels  FB  and  GO,  as  the  square  BA  GF. 
Therefore 


18  ^"*; 


-  ^t 

'•■% 

'  Wm 

•  ] 

m. 

SI 

^ 

^ 

' 

"Sr 

i 

fii 

\r^ 

r      »iti 

•!  wi 

lifcaHi 

3.  Because  ^Z-  is  (by  construction)  parallel  to  BD,  the 
triangle  ABD  is  upon  the  same  base  BD,  and  between  the 


144 


BOOK  IV,     OF  ABBA& 


same  jarallels  BD  and  ^  A  as  the  rectangle  BL,    Therefore 

Area  BL  =  %  area  ABB. 
\       4.  Comparing  (?)  and  (3)  with  (1), 
I  Area  BA  QF  =  area  BL, 

'       5.  In  the  same  way,  by  joining  B£:  and  AK  it  may  be 
shown  that  "^ 

Area  A  QKH  =  area  CL, 
6.  Adding  (4)  and  (5), 

Area  {,BA  OF-\-A  OKH)  =  area  (JBL  +  CL) 

=  square  BCED,  Q.E.D. 
309.  Scholium.  This  proposition  is  called  the  Pythago- 
rean proposition,  because  it  is  said  to  have  been  discovered  by 
Pythagoras,  who  sacrificed  a  hecatomb  of  oxen  in  gratitude 
for  so  great  a  discovery.  It  is  one  of  the  most  important 
propositions  in  geometry,  as  upon  it  is  founded  a  great  part 
of  the  science  of  measurement.  It  also  furnishes  the  basis  of 
trigonometry. 

Corollary.  An  important  special  case  of  this  problem 
occurs  when  the  two  sides  of 
the  triangle  are  equal,  or  when 
AB  =z  AC.  Since  the  squares  on 
AB  and  AC  are  then  equal,  we 
have 

BC  =  AB'  +  AC'  =  2AB\ 
If  we  complete  the  square  by 
drawing  BD  parallel  to  AC  and 
CD  parallel  to  AB,  A  BCD  will 
be  a  square,  and  BC  its  diagonal.     Hence: 

310.  The  square  on  the  diagonal  of  a  square  is  double  the 
square  itself. 

Theorem  XL 

311.  If  from  the  right  angle  of  a  right-angled 
triangle  a  perpendicular  he  dropped  upon  the  hy- 
pothenuse  the  square  of  this  perpendicular  will  be 
e^r^^^v  vu  v,ov  icvvwivyiG  oj  me  iwo  parts  of  the  hv- 
pothenuse,  ^  . 


\ 


A^BA8  OF  PLANE  FIGURES, 


146 


Hypothesis.     ABC,  a  triangle,  right^gled  at  A;  AD,  a 
perpendicular  from  A  on  BC, 

Conclusion.  AD"  =  BD.DG. 

Proof.    1.  Because  BAD  and 

CAD  are  both  right-angled  at  i), 

AB^  -  BD^  =  AD\  X  .g_._. 

AC^  -  DC'  =  AD\\  ^»^^^^ 

2.  Adding  these  two  equations, 

AB'  -i-AC-  BD'  J  DC  =  2AD' 

3.  Because  BA  C  is  right-angled  at  A 

AB' -i- AC  =  BC\  ' 

4.  Comparing  (2)  and  (3), 

BC  -  BD'  -  DC  =  2AD' 
6.  Because  the  line  BCis  the  sum  of  the  lines  BD  and  DC 

BC-BD^-DC^=2BD.Da  (8  289) 

6.  Comparing  (4)  and  (5),  ^®     ^ 

2AD'  =  2BD  .  DC, 


(§  308) 


or 


AD'  =  BD.DC.    Q.E.D. 


Theorem  XII. 
313.  The  square  on  a  side  opposite  any  acute 
angle  of  a  triangle  is  less  than  the  sum  of  the  squares 
on  the  other  two  sides  hy  twicetJie  rectangle  contained 
by  either  of  those  sides  and  the  projection  of  the  other 
side  upon  it. 

Hypothesis.  ABC,  any  triangle  having  the  angle  at  A 
acute;    CD,    the  perpendicular  ^ 

dropped  from  C  on  D,  and  there- 
fore 

AD  the  projection oiACon  AB. 
Conclusion. 

BC^  =  AC  +  AB^-2AB.AD. 
Proof.    1.  Because   CDB  is 
right-angled  at  D,  j__ 

2.  Because  A  CD  is  right-angled  at  D, 
CD'  =  AC-AD\ 


"i"  '11  i 

■^   y-'M  1 

146 


BOOK  IV.     OF  AREAS. 


3.  Putting  this  value  of  CD*  in  (1), 

BC  :=z  BD' -  AD' -{.  AG\ 

4.  BD'  -  AD'  =  (BD  +  AD)  (BD  -  AD), 

=  AB  (BD  -  AD), 
=  AB  (BD  +  AD-  2AD), 
=  AB  {AB  -  2AD), 
=  AB'-^2AB,AD. 
6.  Substituting  this  last  value  in  (3), 

BC  =  AO'-\-AB'-  2AB .  AD, 
Theorem  XIII. 


(§  293) 
(§  287) 


(§  287) 
Q.E.D. 


313.  In  an  obtuse-angled  triangle  the  square  on 
the  side  opposite  the  obtuse  angle  is  greater  than  the 
sum  of  the  squares  on  the  other  two  sides  by  twice  the 
rectangle  contained  by  either  of  those  sides  and  the 
projection  of  the  other  side  upon  it. 

Hypothesis.  ABC,  a  triangle,  obtuse-angled  at  ^;  CD,  the 
perpendicular  from  C  upon  AB  q 
produced,  so  that  DA  is  the  pro- 
jection of  CA  on  AB. 

Conclusion. 
BC  =  AC  -{-  AB'  +  2AB.AD. 

Proof.    1.  Because  CDB  is 
right-anglp'l  at  D, 

BC  =  BD'  4-  CD'.  JD A ^B 

2.  Because  CD  A  is  right-angled  at  D, 

CD'  =  AC  -  AD'. 

3.  Putting  this  value  of  CD'  in  (1), 

BC  =  AC -\- BD' -  AD'. 

4.  BD'  -  AD'  =  (BD  -  AD)  {BD  -f  AD),      (8  293) 

=  AB  {BD  +  AD), 

=  AB{AB-\-2AD), 

=  AB'  +  2AB .  AD.  (§  287) 

5.  Substituting  this  last  value  in  (3), 

BC  =  AC'  +  AB'-i-2AB.AD.    Q.E.D. 

314.  Scholium.    The  method  of  demoiistration  is  the 
same  in  the  last  two  problems,  except  that  in  Th.  XII.  the 


AREAS  GF  PLANE  FIGURES. 


147 


line  AB  is  the  stim  of  the  lines  AD  and  BD,  and  in  Th  XIII 
It  is  equal  to  their  difference.  But  if  we  regard  the  projecl 
tion  AD  as  algebraically  negative  when  it  falls  outside  the 
triangle,  as  in  Th.  .XIII.,  then  Th.  XII.  will  express  both 
theorems,  because  the  subtraction  of  the  negative  rectangle 
AB .  AD  would  mean  that  it  was  to  be  added  arithmetically. 

Theorem  XIV. 

315.  The  projections  of  a  straigJU   line   upon 
parallel  straight  lines  are  equal. 

Proof.   Let  AB  be  the  line  projected,  and  MN,  PQ,  its 
projections   upon    two    parallel  ^b 

lines.  *-  ^^ 

1.  Because    these   lines   are 
parallel,  the  perpendiculars  AM  At 
and  AP,  BQ  and  QN,  form  two  I                               j 
straight  lines.                                   ~jL 1 

2.  Because  the  lines  PJfand  gjV^are  perpendicular  to  the 
same  straight  line  MN, 

MP  II  NQ. 

3.  Therefore  MNPQ  is  a  parallelogram,  and 

MN=:PQ{^m),     Q.E.D. 

Theorem  XV. 

316.  The  sum  of  the  squares  upon  the  two  diag- 
onals of  a  parallelogram  is  equal  to  the  sum  of  the 
squares  upon  the  four  sides.  ^ 

Proof.   Let  ^5CZ)  be  the  paral-         /^ 

lelogram,  having  an  acute  angle  at       /      "^^^ 
A,    Then—  ^  ^ 

1.  In  the  triangle  ^^(7,  A^ ^B 

BC^  =  AB^  +  AC^-2ABx  proj.  of  ^ C on  ^^.     (8 312) 
AD'==  AO^  +  CD^J^^OD  X  proj.  of  AC  on  CD.     (1 313 

^.  Because  AB  and  CD  are  parallel, 

Proj.  of  AC  on  CD  =  proj.  of  ^  O'  on  AB.       (8  315) 
Also,  becansfi  4  nCIT)  {a  p  T.o,«,n^i^ — „_,  ^°        ■' 

AB  =  CD, 
Therefore  the  laat  two  terms  of  the  equations  (1)  are  equal. 


\ 

in 

r'i 


i     ! 


li.,.ilji[ 


148 


BOOK  IV.     OF  AREAS. 


3.  Adding  the  equations  (1),  tlie  last  terms  of  the  equa- 
tions (1)  cancel  each  other,  and  wo  have 
BC  -{■AD'  =  AB'-\-A C  +  A 6"-f  CD\ 

=  AB""  +  BD"  +  CD'  +  A  C"(because  A  C^BD), 
That  is,  the  sum  of  the  squares  on  the  diagonals  AD  and  BC 
is  equal  to  the  sum  of  the  squares  upon  the  four  sides.   Q.E.D. 


«  »  • 


CHAPTER    111. 

PROBLEMS    IN    AREAS 


Problem  I. 


317.  To  construct  a  triangle  which  shall  he  equal 
in  area  to  a  gimn  polygon. 

Given.   A  polygon,  ABCDEF. 

Required.    To  construct  a  triangle  equal  to  it  in  area. 

Construction.   1.  Join  the  ends  of  any  pair  of  adjacent 
sides,  say  BC  and  Ci>,  by  the  line 
DB. 

%  Through  the  intermediate 
angle  C  draw  a  line  parallel  to  BD, 
meeting  AB  produced  in  B'.  F< 

3.  The  polygon  AB'DEF  will 
have  the  number  of  its  sides  one 
less  than  ABCDEF,  because  the 
two  sides  BC,  CD  are  replaced  by 
the  one  side  B'D,  and  it  will  be  equal  in  area. 

4.  By  performing  the  same  operation  upon  AB'DEF,  the 
number  of  sides  will  be  still  further  diminished  by  one,  and 
the  operation  may  be  repeated  until  the  number  of  sides  is 
reduced  to  three. 

Proof.  1.  Because  the  triangles  DCB  and  DB'B  are  on 
the  same  base,  DB,  and  have  their  vertices  on  the  line  CB^ 
parallel  to  that  base,  they  are  equal  in  area  (§  302). 


PROBLEMS. 


149 


2.  The  original  polygon  is  made  up  of  the  parte 
Area  A  BDEF  -f  area  D  CB, 
and  the  new  one,  AB'DEF,  is  made  up  of 
Area  ABDEF+  area  BB'D. 
8.  Because  the  area  DCB  =  BB'Dy 

Area  AB'DEF  =  area  ABCDEF. 
4.  In  the  same  way  it  may  be  shown  that  each  transformed 
polygon  IS  equal  to  the  one  from  which  it  is  formed,  so  that 
the  last  one  of  aU,  which  is  a  triangle,  is  equal  in  area  to  the 
original  polygon. 

Problem  II. 

318.  To  describe  a  parallelogram  which  ^hall  he 
equal  in  area  to  a  given  triangle,  and  ham  one  of  its 
angles  equal  to  a  given  angle. 

Given.   A  triangle,  ABC]  an  angle,  X. 

Required.   To  construct  a  parallelogram  having  one  of  its 
angles  equal  to  X,  and  its 
area   equal   to   the   area 
ABO. 

Construction.  1.  Bisect 
the  base  AB  ot  the  tri- 
angle at  the  point  D. 

2.  Through  C  draw  a 
line  Ci^  parallel  to  AB.     ^i 

3.  At  D  make  the  an- 
gle BDE  =  X,  and  continue  the  side  until  it  meets  OF  in  E 

4.  Through  B  draw  BF  parallel  to  DE.  DBFE  will  then 
be  the  parallelogram  required. 

The  proof  is  left  as  an  exercise  for  the  student. 

319.  Corollary.  If  it  be  required  to  construct  a  rect- 
angle which  shall  be  equal  to  the  triangle,  we  have  only  to 
make  the  Ime  DE  perpendicular  to  AB. 

Problem  III. 

330.  To  describe  a  smm.rp.  onMnJi  0*^77  7..,  ^ 7  .•„ 

area  to  a  given  rectangle. 


^-V' 

TS 

t 

\ 

-  ^ 

■    ■ 

1  ' 

^  |i 

■  f 


160 


BOOK  IV.     OF  AJiEAS. 


I  ! 


!     I 


m 


1 


D 


Given.   A  rectangle,  A  BCD. 

Required.   To  construct  a  square  of  the  same  area. 

Analysis.  Theorem  X.  teaches  that  in  the  right-angled 
triangle  ABGy  the  square  upon  AD  is  equal  to  the  rectangle 
BD  .  DC. 

Therefore,  if  we  can  construct  a  right-angled  triangle  in 
which  the  perpendicular  from  the  right  angle  upon  the  hy- 
pothenuse  shall  divide  the  latter  into  two  parts  equal  respect- 
ively to  two  adjacent  sides  of  the  given  rectangle,  this  perpen- 
dicular will  be  the  side  of  the  square  required. 

This  result  will  be  reached  by  describing  a  semicircle  upon 
a  diameter  equal  to  the  sum  of  two  adjacent  sides  of  the  rect- 
angle, because  all  the  angles  in  the  semicircle  are  right  angles. 

Construction.  1.  Produce  AB  to  the  point  P,  making 
BP  =  BG. 

2.  On  AP  as  a  di-  -^- JL 

ameter  describe  a  semi- 
circle. 

3.  Produce  BC  until 
it  cuts  the  semicircle 
in  Q. 

The  square  on  BQ 
will  be  the  square  ..'equired. 

Proof.  1.  If  we  join  AQ,  PQ,  the  triangle  AQP  will  be 
right-angled,  because  it  is  inscribed  in  a  semicircle. 

3.  Because  BQiaa.  perpendicular  from  the  right  angle  Q 
upon  the  base  AP,  we  have 

BQ'  =  AB.BP  =  AB.BC.    Q.E.D. 

331.  Corollary.  By  the  three  preceding  constructions  a 
square  may  be  constructed  equal  in  area  to  any  given  polygon. 
The  polygon  is  first  transformed  into  a  triangle  by  Problem  I. ; 
this  triangle  into  a  rectangle  by  Problem  II.,  Cor.;  this  rect- 
angle into  a  square  by  Problem  III. 

Peoblem  IV. 

322,  To  describe  a  rectangle  wMcTi  shall  he  equal 
in  area  to  a  gicen  pctrallelogTam. 
Given.    A  parallelogram,  A  BCD. 


0 

N 

\ 
\ 

/ 
/ 
/ 

/ 
/ 

1 
1 

L' 

\ 

P 

I 


l4-.~^i^;| 


PROBLEMS. 


161 


Required.    To  describe  a  rectangle  having  the  same  area. 

Construction.      Produce 
the  side  CD  to  F,  and  at  A   ' 
and  B  erect  perpendiculars 
to  AjB,  meeting  £!F  in  B 
and  F. 

ABFF  will  be  the  rect- 
angle required. 

Proof.    The  proof  is  given  by  Theorem  V. 

333.  Corollary.  If,  instead  of  a  rectangle,  we  wish  to 
describe  a  parallelogram  having  a  given  angle,  we  have  only 
to  make  the  angle  BAF  equal  to  the  given  angle. 


Peoblem  v. 

324.  On  a  gwen  line  as  a  base  to  describe  a 
paraUelogram  equal  to  a  given  parallelogram  in  area 
ana  tn  angles. 

Given.  A  parallelo- 
gram, ABCD;  a  line, 
BM. 

Required.      To    de- 
scribe on  BM9,  parallel- 
ogram having  the  same 
area  as  ABCB  and  equi-  / 
angular  to  it.  * 

b-.o^Tvf'!f' V^'*  5Jf  be  drawn  in  the  same  straight 
Ime  with  AB.    Then—  ^ 

duced  ki^iT^^  ^^''^'^  ^-^^ parallel  to  BC,  meeting  Z^C'pro- 

2.  Draw  the  diagonal  NB,  and  produce  it  until  it  meets 
BA  produced  in  P. 

3.  Through  P  draw  PQR  parallel  to  AM,  and  meeting 
CB  produced  m  Q  and  WM  produced  in  R. 

pij-      .,  \     "^"'-  xcvjuixcii  piiiuhuiogram,  having  Oii;  = 

i^il/  as  Its  base,  and  equal  to  ABCD  in  area  and  in  its  andes. 
Proof.     From  Theorem  IX. 


162 


BOOK  IV.     OF  AHEAS. 


!l    i 


Problem  VI. 

.  325.  On  the  base  of  a  given  triangle  to  describe 
another  triangle  equal  in  area  and  hamng  a  gitm 
angce,  q  t\ 

Given.  A  triangle,^  J5C; 
an  angle,  0. 

Required.  On  the  base 
AB  to  describe  a  triangle 
equal  to  ABC  m  area,  and 
having  an  angle  equal  to  0. 

Construction.  a* 

1.  Through  C  draw  CD  parallel  to  AB. 

3.  At  A  make  the  angle  BAD  =  0,  and  produce  the  side 
until  it  meets  CD  in  D.  '         f 

3.  Join  BD. 

ABD  will  be  the  triangle  required. 

Proof.    From  Theorem  VII.,  Cor.  1. 

Problem  VII. 
326.  To  form  a  triangle  equal  in  area  to  a  given 
triangle,  and  having  its  base  on  the  same  straight 
line  and  its  vertex  in  a 
given  point. 

Given.  A  triangle,  ABC\ 
a  points  P' 

Required.  To  describe  a 
triangle  equal  to  ABC  in  area, 
having  its  base  on  AB  and  its 
vertex  in  P. 

Construction,  1.  Join^P 
and  PB. 

3.  Through   C  draw    CD^ 
parallel  to  AB,  meeting  AP  in  D. 

3.  Draw  DQ  parallel  to  PB,  meeting  AB  in  0. 

4.  Join  Pq.  6  V 

AQr  will  be  the  required  triangle  on  the  base  AB\  equal 
m  area  to  ABC,  and  having  its  vertex  in  P, 


I 


H--^M 


PROBLEMS.  153 

Proof.  Join  BD.    Then— 

1.  Because  AB  and  CD  are  parallel,  ' 

Area  ABC—  area  ABD. 
3.  Becaust)  D(^  and  P5  are  parallel, 

Area  BPB  =  area  ^P^. 

3.  Area  ABD  +  BPD  -  BPQ  =  area  A  QP, 

4.  Comparing  with  (2), 

Area  ABD  =  area  AQF. 

5.  Comparing  this  with  (1), 

Area  Jl^C7  =  area  ^^P.    Q.E.D. 

The  construction  and  demonstration  of  the  following 
problems  are  left  as  exercises  for  the  student. 

Problem  VIII. 

327.  To  construct  a  square  which  shall  be  equal 
to  the  sum  of  two  given  squares. 

Eemark.  If  we  form  a  right-angled  triangle  of  which 
the  sides  about  the  right  angle  are  equal  to  the  sides  of  the 
given  squares,  the  square  upon  the  hypothenuse  will  be  that 
required  (§  308). 

Peoblem  IX. 

328.  To  construct  a  square  which  shall  be  equal 
to  the  difference  of  two  given  squares. 

Peoblem  X. 

329.  To  construct  a  square  which  shall  be  equal 
to  one  half  ^  gi'aen  square. 

Peoblem  XI. 

330.  To  divide  a  triangle  into 
any  given   number  of  equal    tri- 


annlfis 

i7 


■if 


I  linn  a  o 


vertex  to  the  base. 


WW 


:f!lfi 


p 

1^ 

It 

\ 

■r 
< 

i 

'1 

■ 

It 

.      t 

M 

t 

1 

i 

1 

.,ilii 

11 

I  i 


I 


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Nii 


154 


BOOK  IV.     OF  ABEA8. 


CHAPTER  IV. 

THE  COMPUTATION   OF  AREAS. 


331.  Geometrical  problems  may  be  divided  into  two  gen- 
eral classes,  depending  on  the  kind  of  solution  which  is  to  be 
obtained. 

^  I.  Problems  of  pure  geometry.  In  problems  of  construe- 
tion  the  solution  consists  in  drawing  a  figure  which  is  to  con- 
form to  the  conditions  of  the  problem.  The  answer  to  such 
a  problem  is  given,  not  in  numbers  or  algebraic  expressions 
but  as  a  geometrical  figure  simply.  The  problems  we  have 
hitherto  considered  belong  to  this  class,  and  they  are  the  only 
kind  recognized  as  belonging  to  pure  geometry. 

II.  Problems  of  numerical  geometry.  In  problems  of  the 
second  class  the  solution  appears  not  merely  as  a  line  or 
figure  drawn  upon  a  plane,  but  as  a  calculated  length  or  a 
calcula^  3d  extent  of  area.  For  example,  the  result  may  be 
expressed  by  saying  that  a  line  the  length  of  which  is  re- 
quired is  seven  centimeters  or  other  units  in  length,  or  that 
a  surface  contains  a  certain  number  of  square  units  The 
number  of  units  in  either  case  may  be  expressed "  either 
by  algebraic  symbols  or  by  the  numbers  of  arithmetic.  Such 
problems  may  be  considered  as  belonging  to  numerical  or 
algebraic  geometry. 

Relatims  of  the  ttvo  methods.  In  pure  geometry  the 
division  of  magnitude  into  definite  units  is  not  recognized. 
If  an  angle  is  given,  it  is  supposed  to  be  given  by  drawing  it 
not  by  stating  the  number  of  degrees.  The  angle  itself  may 
not  be  drawn  at  all  except  in  imagination.  So,  also,  a  given 
length  is  a  length  of  a  given  line,  and  not  a  number  of  units 
of  any  kind. 

Pure  geometry  was  almost  the  only  kind  cultivated  by  the 
ancients,  because  the  methods  of  algebra  were  not  known  to 


COMPUTATION  OF  AREAS. 


166 


them.  Hence  it  is  sometimes  called  the  ancient  geometry. 
But  it  does  not  suffice  for  modern  wants,  where  numbers  of 
miles,  feet,  acres,  etc.,  are  required. 

One  great  advantage  of  the  modern  method  arises  from 
the  application  of  algebraic  signs  to  lines.  In  the  ancient 
geometry,  whenever  the  position  of  a  point  is  changed  to  the 
opposite  side  of  a  line,  we  have  to  suppose  a  different  theorem 
or  a  different  case  of  the  same  proposition.  But  in  the  modern 
geometry  the  difference  is  expressed  by  changing  the  algebraic 
sign  of  the  distance  of  the  point  from  the  line,  and  the  general 
statement  of  the  proposition  remains  the  same. 

The  general  investigation  of  lines  and  areas  by  algebraic 
methods  requires  an  application  of  trigonometry,  and,  in  the 
cases  of  curve  lines,  of  the  integral  calculus;  but  there  is  a 
general  method  applicable  to  the  computation  of  areas  both 
m  the  surveying  of  land  and  in  the  integral  calculus,  the 
principles  of  which  can  now  be  explained. 

Problem  XII. 

332.  To  find  hy  measurement  and  calculation  the 
area  of  a  given  polygon.         n 

Let  ABODE  be  the  polygon.  C 

Draw  any  straight  line  MN. 
From  each  angle  of  the  polygon 
drop  a  perpendicular  upon  the  ^ 
line  MN.     Let  A',  B',  C\  etc.,  B' 
be    the   points   at  which  these 
perpendiculars  meet  the  line. 

The  area  ^M^CC"  includes  E' 
the  whole  polygon  plus  an  area 
AEDCC'A'  between  the  poly- 
gon and  the  line.  Therefore, 
if  from  the  first  of  these  areas 
we  take  away  the  second,  the 
remainder  will  be  the  area  of 
the  polvffon. 

Each  of  these  two  areas  is  made  up  of  several  trapezoids 
namely:  ^         ' 


il 


|£      tt 

li. 

\  1 


166 


BOOK  IV.     OF  AREAS. 


First  area  =  trapezoid  A' ABB' 

+  trapezoid  B'BCC, 

Second  area  =  trapezoid  A' ABE' 

+  trapezoid  E'EDD* 

+  trapezoid  D'DCC, 

The  first  area  includes  the  quantities  to  be  added,  the 

second  those  to  be  subtracted. 

The  area  of  each  trapezoid  is  the  rectangle  of  its  altitude 
into  half  the  sum  of  its  parallel  sides  (Th.  VIII.).  In  par- 
ticular. 

Area  A' ABB'  =  J  (A' A  +  B'B)  A'B\ 

Area  B'BCC  =  ^  (B'B  +  CO)  B'C. 

Area  O'CDD'  =  ^  (C'C  +  D'D)  G*D\ 

etc.  etc.  etc. 

To  express  these  areas  in  algebraic  form  let  us  put  j3i,  p%y 

pz,  etc.,  for  th6  lengths  of  the  several  perpendiculars;  that  is, 

p\  =  A' A, 
P'i  =  B'B, 
pz=  CO, 
p*  =  D'D. 
ps  =  E'E. 
Let  us  also  take  an  arbitrary  point  0  on  the  line,  to 
measure  distances  from,  and  put 

yi  =  0A\ 
y^  =  0B\ 
yz  =  0C\ 
^4  =  OD'. 
yr>  -  OW. 
Then  A'B'  =  yi-  yu 

B'C  =  y3-  yu 
CD'  =  2^3  -  yu 
D'E'  =  yi-  y5. 
E'A'  =  ys  -  yu 
The  expressions  for  the  areas  vvIU  then  be: 

Area  A' ABB'  =  |  {r-,...  f  p^)  {^^  _  y^y 
B'BCC  =z}^{p,^p,)(y,-y,y 
OG'Djy  =  ^  (j04  +  jt73)  {yz  -  y,). 
n'JJEE'  =  i  ( j05  +^4)  («/4  -  y,). 
E'EAA'  =^{p,J^p,)(y,^^y{), 


! 


COMPUTATION  OF  AREAS. 


167 


Ihe  required  area  of  the  polygon  we  have  found  to  be 
given  by  subtracting  the  last  three  areas  from  the  first  two 
Now  this  subtraction  may  be  indicated  by  simply  changing 
the  algebraic   signs  of  the  quantities  to  be  subtracted— a 
change  which  will  be  effected  by  changing  the  factor 

^8  —  jji  into  ^4  —  yz. 
y^  —  yi  into  yt,  —  yu 
yi  —  y\  into  y\  —  y^. 
The  expression  for  the  area  will  then  be: 
Area  ABODE  =  i  {p^  +  jp.)  {y^  -  y,) 
+  Hp3-\-]h)  (yi-yi) 
+  i(^4+i»3)  (y*  -yz) 

-{-^{pi  +j06)  {yi  ~y,). 

It  will  be  seen  that  the  formula  is  uniform  with  respect 
to  the  different  values  of  p  and  y  taken  in  order,  each  value 
of  p  being  added  to  that  next  following  in  order,  and  each 
value  of  y  subtracted  from  that  next  following  in  order. 

If  we  execute  the  multiplications  indicated,  one  half 
the  partial  products  will  cancel  each  other,  and  the  area  will 
reduce  to 

ilpiy^  —  p^yx 
•j-piya  —  piyi 
-\-  pzyi  —  p4yz 
-{-piys  —  p'\y4 
-{-p^yx  —pxys] 
In  principle  this  method  is  that  used  by  surveyors  in 
computing  the  area  of  irregular  pieces  of  land.     It  also  in- 
volves the  best  system  of  measuring  areas  in  more  advanced 
mathematical  investigations. 

The  student  may  be  supposed  to  find  the  values  of  pi,  p^, 
etc.,  yx,  yi,  etc.,  by  measurement  on  the  actual  figure.  Their 
calculation,  when  all  the  sides  and  angles  are  given,  is  a  prob- 
lem of  trigonometry. 

The  criterion  whether  a  trapezoid  is  to  be  put  into  the 
additive  or  the  subtractive  column  is  this: 

If,  in  crossing  over  anvsiflpnf  flio  T^nlTTn•/^«  ^»^/^r«  +v,<^  ^„4-r,:Ar. 
to  the  inside,  we  pass  to  the  inside  of  the  trapezoid  bounded  by 
that  side,  the  area  of  that  trapezoid  is  additive,  or  aig(  orai- 
cally  positive. 


} 


r 

.1 


,1 1 


i;.' 


I 


I    :i       i 


168 


BOOK  IV.     OF  AREAS. 


If,  in  passing  inside  of  the  polygon,  we  pass  out  of  the 
trapezoid,  the  area  of  that  trapezoid  is  subtractive,  or  alge- 
braically negative. 

Examples.  If  we  pass  into  the  polygon  over  the  side  AB, 
We  pass  into  the  trapezoid  A'ABB*.  Therefore  the  area  ol 
this  trapezoid  is  additive.     (See  diagram  on  p.  155.) 

The  same  applies  to  B'BCC. 

If  we  pass  into  the  polygon  over  the  side  CD,  we  pass  out 
of  the  trapezoid  C'CDD\  Therefore  the  area  of  this  trape- 
zoid is  negative. 

The  same  remark  applies  to  the  trapezoid  bounded  by  DE 
and  by  EA. 

EXERCISES. 

Measure  off  and  compute  the  area  of  each  of  the  following 
polygons  in  square  centimetres,  inches,  or  other  scale  measure. 


It  will  be  noticed  that  owing  to  the  re-entrant  angles  of  the  second 
figure  there  is  a  double  overlapping  of  some  of  the  trapezoids.  But 
this  makes  no  change  in  the  application  of  the  formula,  which  always 
gives  correct  results  when  the  algebraic  signs  of  the  quantities  arc 
properly  interpreted. 

333.  Algebraic  expression  for  the  Area  of  a  Triangle. 
Because  a  triangle  is  completely  determined  when  three  ol 
its  sides  are  given  (§  110),  its  area  must  admit  of  being  ex- 
pressed algebraical^  in  terms  of  its  sides.  The  required  ex- 
pression is  found  as  follows: 


AREA  OF  A   TRIANGLE. 


159 


Let  ABC  be  the  triangle,  and  CD  the  perpendicular  from 
C  upon  D.     Put  ; 

a,  the  side  BC; 
by   "     "    AC', 


(t 


t( 


AB; 


JO,  the  perpendicular  CD, 
Then 

Area  ABC  X  2  =  cp.  (§  301) 
To  find  p  we  have,  from  the     /         , 

right-angled  triangles   CDA  andjj- -J- 

CDB,  » 

p^  =  AC'-  AD'  =  BC  -  BD'  =  BC'  -  (AB  -  ADY 
or  \  /  > 

f  =  h'-  AD'  =  a'-{c-  ADy  =  a'-c'  +  2c.AD-AD\ 
We  must  use  these  equations  to  eliminate  the  quantity  iD 
from  the  expression  for^.  Equating  the  second  and  fourth 
members  of  the  last  line  of  equations,  we  find 

*'  =  a'  -  c"  +  2c.AD: 
whence 

2c 
Substituting  the  square  of  this  in  the  expression  for  p\  we 

4c»  —435  -. 

Squaring  the  above  pxpression  for  the  area, 
4(Area^^c    '  =  cy. 

16(Area  ABCy  =  4c>'  =  43V»  -  (5»  +  c'  -  ««)» 
This  expression,  being  the  difference  of  two  squares,  may  be 
transformed  into  the  product  »      j  "« 

(2bc  -\-b'-i~c'-  a')  (2bc  -  b'-c'-{-  an 
The  first  three  terms  in  the  first  factor  are  a  perfect  square- 
namely,  the  square  of  b  +  c;  and  the  first  three  of  the  second 
factor  are  the  negative  of  the  same  square.     Therefore  each 
lactor  can  again  be  factored,  making  the  product 

{b-^^c  +  a){b  +  c~a){a-i-b-c)(a~b  +  c), 
s  =  U^ -\- b -{- c)  or  28  =  a -\-b  +  e, 


i  ; 


m 


i'! 
1 1 ' 


:!!  :i 


160 
wo  have 


BOOK  IV.     OF  AHEAa. 


b  -\- c  -\-  a  =  'Zs. 

b-^c  —  a  =  2(v  -  a). 

a-\-i  —  c  ~    ;^o  —  v), 

a  —  b  -\-c  =  2{s  —  b). 

Substituting  these  values  and  dividing  by  16,  wo  have 

(Area  ABOy  =  s{s-  a)  {s  -  b)  (a  -  c) 
and 

Area  ABC  =   Vs{s  -  a)  (s  -  b)  (s'^c), 
the  required  expression. 

TlIEORKMS   FOR  EXERCISE. 

Theorem  1.  The  dififerencc  of  the  squares  upon  any  two 
sides  of  a  triangle  is  equal  to  the  difference  of  the  squares  of 
the  projections  of  these  sides  upon  the  third  side. 

Theokem  2.  The  sum  of  the  squares  upon  the  diagonals 
of  a  quadrilateral  is  equal  to  twice  the  sum  of  the  squares 
upon  the  four  lines  joming  the  middle  points  of  its  sides, 
taken  consecutively.     (See  Th.  21,  p.  89.) 

Theorem  3.  If  we  join  the 
middle  points  of  two  opposite 
sides  of  a  quadrilateral  to  two 
opposite  angles,  the  two  triangles 
thus  formed  will  have  half  the  area 
of  the  quadrilateral. 

HypotJmis.    FG  =FD,  AE=  EB. 

Conclusion.    Area  AFB  +  EBG  =  area  AECF 

=  i  area  ABGB. 

Theorem  4.  The  four  triangles  into  which  a  parallelo- 
gram is  divided  by  its  diagonals  are  of  equal  area. 

Theorem  5.  If  from  any  point  on  the  diagonal  of  a 
parallelogram  lines  be  drawn 
to  the  opposite  angles,  the 
parallelogram  will  be  divided 
into  two  pairs  of  equal  tri- 
angles. 

Area  OAD  =  area  OAB. 

Area  OCB  =  area  OGB. 


K-i^ 


B 


mEOUEMS  FOR  EXER0I8E.  igj 

Theorem  6.  Tin-  parallelogram  formed  by  joining  the 
middle  points  of  the  consecutive  sides  of  a  quadrilateral  has 
halt  the  area  of  the  quadrilateral  (§  13G). 

;  Theorem  7.  If  through  the  middle  point  of  one  of  the 
non-parallel  sides  of  a  trapezoid  we  draw  a  line  parallel  to  the 
opposite  side,  and  complete  the  parallelogram,  tlie  area  of  the 
parallelogram  will  be  equal  to  that  of  the  trapezoid. 

Theorem  8.     If  we  join  the  middle  of  one  of  the  non- 
parallel  sides  of  a  trapezoid  to  the  ends  of  the  opposite  side 
the  middle  triangle  will   lave  half  the  area  of  the  trapezoid.   ' 

Theorem  9.  If  two  triangles  have  two  sides  of  the  one 
equal  to  two  sides 

of  the  other   re-  q 

spectively,  and 
the  included  an- 
gles supplemen- 
tary, they  are 
equal  in  area. 

Hypothesis.     GA  =  MK.     GB  =  ML. 

Angle  AGB-\-  angle  KML  =  180°. 

Condumn.    Area  ABG  =  area  KLM. 

Theorem  10.  The  sum  of  the  squares  upon  the  diagonals 
of  a  trapezoid  are  equal  to 
the  sum  of  the  squares  upon 
the  non-parallel  sides  plus 
twice  the  rectangle  of  the 
parallel  sides. 

Gonclumn.    AG^ -\-  BBP  -  AD"^  -f  5C«  +  2AB.  GD. 

Theorem  11.  If  from  any 
point  within  a  polygon  perpen- 
diculars be  dropped  upon  the 
sides,  the  sum  of  the  squares  of 
one  set  of  alternate  segments  is 
equal  to  the  sum  of  the  squares 
of  the  other  set. 

Aa?  -I-  Bb^  +  Gc^  -f  Bd^  +  Ee^  =  aB^  +  hG^  +  cL'^  +  ~aE^i  ^  ,^.. 


''I 

51 


■  i 


;if 


x 
lif 


!l 


::*••■ 


.      1 

i 

1 

i 
,   i 

^B 

1  ^H 

'^ra^l 

iili 


ill 


i 


162 


BOOK  IV.     OF  AUEA8. 


Numerical  Exercises. 


1.  In  a  right-unglod  triangle  the  lengths  of  the  sides  oon- 
tttiniug  the  riglit  angle  are  9  and  12  feet.  What  is  the  length 
of  the  hyiwtheniiso?    What  is  the  area  of  the  triangle? 

2.  If  the  length  of  the  hypothenuse  is  10  feet,  and  that  of 
one  side  8  feet,  wliat  is  the  length  of  the  remaining  side? 
What  is  the  area  of  tlio  triangle? 

3.  In  a  riglit-angled  triangle  the  perpendicular  from  the 
right  angle  upon  the  hypothenuse  divides  the  latter  into 
segments  wliich  are  respectively  9  and  10  feet.  Find  the 
lengths  of  the  perpendicular  and  of  the  two  sides,  and  the 
area  of  the  triangle. 

4.  What  three  different  expressions  for  the  area  of  a 
triangle  may  we  obtain  from  §  301  by  taking  different  sides 
as  the  base?    What  theorem  hence  follows? 

5.  What  is  the  area  of  the  triangle  of  which  the  respective 
sides  are  15,  41,  and  52  metres? 

6.  If  the  diagonal  of  a  rectangle  is  13  feet,  and  one  of  the 
sides  12  feet,  what  is  the  area? 

7.  Show  how  the  altitude  and  area  of  a  trapezoid  may  be 
computed  when  its  four  sides  are  known. 

Refer  to  the  computation  of  the  altitude  p  of  a  triangle  in  §  833. 

8.  If  each  side  of  an  equilateral  triangle  is  unity,  find  its 
altitude. 

9.  Draw  an  equilateral  triangle,  ABO. 
Show  that  the  bisectors  of  each  interior 
angle  will  bisect  the  opposite  side  perpen- 
dicularly. Show  that  if  the  bisector  of  0 
be  produced  beyond  the  point  0  in  which 
it  meets  the  other  bisectors  and  intersect 
the  opposite  side  in  D,  and  if  wo  take 
DF=DO  and  join  AF,  BF,  then— 

I.   0^1  i^^  and  O^jP  will  be  equilateral  triangles. 
II.  The  points  A,  C,  B,  F  lie  on  a  circle. 
III.  The  lines  OA,  OB,  and  0(7  will  all  be  equal. 
Also,  supnosinff  the  lenerth  of  each  side  of  the  trianjylp. 
ABO  io  be  unity,  compute  the  lengths  of  OCand  OD. 


BOOK  V. 
THE  PROPORTION  OF  MAGNITUDES, 


CHAPTER    I. 


RATIO    AND    PROPORTION    OF    MAGNITUDES    IN 

GENERAL 


334.  Definition.  When  a  greater  magnitude  con- 
tains a  lesser  one  an  exact  number  of  times,  the  greater 
one  is  said  to  be  a  multiple  of  the  lesser,  and  the 
lesser  is  said  to  measure  the  greater,  and  to  be  an 
aliquot  part  of  the  greater. 

335.  Def.  When  a  lesser  magnitude  can  be  found 
which  is  a  measure  of  each  of  two  greater  ones,  the 
latter  are  said  to  be  commensurablei  and  the  former 
is  said  to  be  a  common  measure  of  them. 

336.  Def.  When  two  magnitudes  have  no  com- 
mon measure  they  are  said  to  be  incommensurable. 

337.  Def.  When  one  of  two  commensurable 
magnitudes  contains  the  common  measure  m  times, 
and  the  other  contains  it  n  times,  they  are  said  to  be 
to  each  other  as  m  to  n. 

Example.     If  the  magni- 
tude A  contains  the  measure 
a  5  times,  and   B  the  same 
measure  3  times,  then  ^  is  to       '         '         '         ' 
^  as  5  to  3,  and  «  is  a  common  measure  of  A  and  B. 

Exercises.  Draw,  by  the  eye,  pairs  of  lines  which  shall 
be  to  each  other  as  3  to  4;  as  2  to  5;  as  4  to  7;  as  5  to  6;  as 
7  to  %. 


A 


a 


B 


a 


a 


"1 

)    1 

1 

i    ' 

n 

/       \ 

\ 


nejl^ 


'[■( 


s/r:MA 


I 


i 


164         BOOK  V.     PROPORTION  OF  MAQNITUDES. 

338,  Corollary,  If  ^  is  to  -6  as  m  to  n,  then,  by  defini- 
tion (§  337),  the  With  part  of  A  will  be  equal  to  the  nth.  part 
of  B',  or,  in  symbohc  language, 

A       B 


A 

5 


A 

5 


A 

5 


A 
5 


A 

5 


m       n 
Note.    The  mth.  part  of  a  mag- 
nitude is  indicated  by  a  fraction  of     L 
which  the  symbol  of  the  magnitude  Whole  magnitude  A. 

is  the  numerator  and  m  the  denominator. 

Notation..  The  statement  that  two  magnitudes  A  and 
B  are  to  eae>i  other  as  the  numbers  m  and  n  is  written 
symbolically 

A  :  B  ::  m  .  n, 
or  A  :  B  =  m  :  n. 

Note.  The  second  form,  or  that  of  an  equation,  is  preferable,  and  is 
most  used  by  mathematicians;  but  the  first  form  is  more  common  in 
elementary  books. 

339.  Def.  If  a  pair  of  magnitudes  A  and  B  are 
to  each  other  as  two  numbers  m  and  n,  and  another 
pair  P  and  Q  are  also  to  each  other  as  m:n,  then  we 
say  that  A  is  to  5  as  P  to  Q,  and  the  four  magnitudes 
A,  B,  P,  and  Q  are  said  to  be  proportional  or  to 
form  a  proportion. 

Notation.  The  statement  that  the  four  magnitudes  A,  B, 
P,  and  Q  are  proportional  is  expressed  in  the  symbolic  form  ' 

A  :  B  ::  F  :  Q, 
or  A  :  B  =  P  :  Q; 

which  is  read:  ^  is  to  ^  as  P  is  to  Q. 

340.  Def.    The    symbolic    statement   that   four 
magnitudes  are  proportional 
is  called  a  proportion. 

Example.     If  A  contains  a 
twice,  and  B  contains  it  three 
times;  if  also  P  contains  jo  twice,  p  _   p       p 
and  Q  contains  it  three  times,      '         '         ' 

then  _  Q  -—P-P p 

A  :  B  ::  P  :  Q.  '  '         '         ' 

341.  I)ef.  The  four  quantities  which  form  a  pro- 
portion are  called  terms  of  the  proportion. 


B 


a 


a 


a 


a 


^1%,^ 


AXIOMS. 


166 


343.  Def.  The  first  and  fourth  terms  of  a  pro- 
portion are  called  the  eictremes;  the  second  and 
third,  the  means. 

Example.  In  the  last  proportion  A  and  Q  are  the  ex- 
tremes, B  and  P  the  means. 

343.  Def.  The  first  and  third  terms,  which  pre- 
cede the  symbol  :  ,  are  called  antecedents;  the  second 
and  fourth,  which  follow  the  symbol  :  ,  are  called 
consequents. 

Example.  In  the  last  proportion  A  and  P  arc  the  ante- 
cedents, B  and  Q  the  consequents. 

344.  Def.  If  the  means  are  equal,  each  of  them 
is  said  to  be  a  mean  proportional  between  the  ex- 
tremes, and  the  three  quantities  are  said  to  be  in  pro- 
portion. 

Axioms, 

345.  Ax.  1.  If  there  be  a  greater  and  a  lesser 
magnitude  of  the  same  kind,  the  greater  may  be 
divided  into  so  many  equal  parts  that  each  part  shall 
be  less  than  the  lesser  magnitude. 

Note.  By  magnitudes  of  the  same  kind  are  meant  those  which  are 
both  numbers,  both  lines,  both  surfaces,  or  both  solids. 

Ax.  2.  If  a  greater  magnitude  be  a  certain  number 
of  times  a  lesser  one,  then  any  multiple  of  that  greater 
one  will  be  the  same  number  of  times  the  correspond- 
ing multiple  of  the  lesser. 

Symbolic  expression  of  this  axiom.  If  mag.  G  =  ix  mag.Z, 

nG=:i  X  nL. 


then 


Ax.  3.  If  a  lesser  magnitude  be  a  certain  aliquot 
part  of  a  greater  one,  then  any  multiple  of  the  lesser 
one  will  be  the  same  aliquot  part  of  the  correspond- 
ing multiple  of  the  greater. 

ic  exnrp-sisi'.nn   n-F  ihi'o  /»..m*/^*v,        t*  , r         mag^.  (? 

-.     .— j_. ,„.,      ,,j       vfi-vv     ltAH///t.  XX.    UlUlHtX/  ^^^    T 


Symbol 


then 


11 


( ( 


^l 


. 


u 


I         'E 


W 


i 


t 


i 


um 


III 


166 


BOOK  V.     PROPOUTION  OF  MAGNITUDES. 


346.  Theorem.  Equimultiples  of  commensurable 
magnitudes  are  'proportional  to  the  nri/agnitudes  them- 
selves. 

Hypothesis.  A  and  B,  two  a w 

commensurable    magnitudes; 

P,   a  magnitude  i  times    as      '  ' 

great  as  A;  Q,  a  magnitude  i  ^'  '' 


times  as  great  as  B,   i  being   Qj ■ 

any  number  whatever. 

Conclusion.  P  :  Q  ::  A  :  B. 

Proof.     1.  Let  the  Ty^r-gnitude  A  he  to  B  asm  to  n. 
This  will  mean  that  if  we  divide  A  into  m  parts  and  B 
into  n  parts,  these  parts  will  be  equal,  or 

A_B 
m~  n* 

2.  Because  P  ~  iA,  if  we  divide  P  into  m  parts,  we  shall 
have  for  each  part 

P       I'A  A     ' 

'Z;  =  ~  =  '^X-'  (§345,  Ax.  3) 

mm  m  \o       ^  / 

3.  In  the  same  way,  if  we  divide  Q  into  n  parts, 

n  n 

4.  Comparing  these  results  with  (1), 

m       n 
Therefore  P  '.  Q  \\  m  \  n.  (§  339) 

Comparing  with  iX),     P  '.  Q  w  A  \  B. 

347.  Corollary  1.  In  a  similar  way  it  may  be  shown  that 
similar  aliquot  parts  of  magnitudes  are  to  each  other  as  the 
magnitudes  themselves. 

That  is,  whatever  be  the  whole  number  it, 

k  '    h  "  ^  '^• 

348.  Cor.  2.  Similar  fractions  of  magnitudes  are  pro- 
portional to  the  magnitudes  themselves. 


}  "J 


y^xrt  wvvx'f 


:/  ^> 


V     iiiC  T  V 


-cP:i:Q::iP:iQ, 


RATIO  OF  TWO  MAGNITUDES.  1^7 

and,  by  the  original  theorem, 

iP  :  iQ  ::  P  :  Q,  ; 

from  which  follows 

•  • 

-]^P''-^Q'-''P''  Q- 

Ratio  of  Two  Magnitudes. 
349.  Consider  any  two  numbers,  which  we  may  call  m 
and  n.     If  we  divide  each  of  them  into  n  parts,  each  part  of 

m  will  be  -,  and  each  part  of  n  will  be  -  =  1.    By  Corollary  1 

of  the  last  theorem  these  parts  will  be  to  each  other  as  the 
original  numbers;  that  is, 


,.f' 


m  :  n 


-  -1 

n 


Therefore,  if  two  magnitudes  A  and  B  are  to  each  other  as  m 

to  n,  they  will  also  be  to  each  other  as-  to  1,  or 

n 


A  :  B 


m 
n 


1. 


350.  Def.    When  two  magnitudes  are  to  each 

/rrt 

Other  as  m  to  n,  the  fraction  ~  is  called  the  ratio  of 

lb 

the  magnitude  A  to  the  magnitude  B. 
Corollary.  When  wo  say 

A  :  B  :-.  m  :  n, 
we  mean  that  A  contains  m  parts,  and  B  contains  n  equal 
parts  (§  338).     Hence: 

351.  The  ratio  of  a  magnitude  A  to  another  magnitude 
B  is  the  quotient  formed  by  dvnding  the  number  of  parts  in 
A  by  the  numher  of  equal  parts  in  B. 

352.  Scholium.  There  are  three  ways  of  conceiving  of 
the  ratio  of  two  magnitudes,  which  all  lead  to  the  same  result. 

I.  If  we  have  two  magni- 
tudes A  and  B,  the  ratio  of  A    a 

to  B  is  the  numerical  factor 

by  which  we   must  multiply   ® 

the  consequent,  B,  in  order  to  produce  the  antecedent,  A. 
There  may  then  be  three  cases: 


A 

n\ 
in 


5i1 


.11 


I 


'I 
'I   L 

I    '\ 

111 


H 


I  1^1 


I'  * 


168         B005'  V.     PROPORTION  OF  MAGNITUDES. 

I.  If  ^  is  a  multiple  of  B,  the  ratio  is  a  whole  number. 
The  multiplication  is  then  effected  by  adding  B  to  itself  the 
proper  number  of  times. 

I        2.  If  ^  and  B  are  commensurable,  the  ratio  is  a  vulgar 
fraction.     If  A  contains  the  common  measure  m  times,  and 

'  B  contains  it  n  times,  the  ratio  is  — .    We  may  conceive  the 

multiplication  to  be  effected  by  dividing  B  into  n  parts,  and 
taking  m  of  these  parts  to  make  A. 

3.  If  ^  and  B  are  incommensurable,  the  ratio  will  neither 
be  a  whole  number  nor  a  vulgar  fraction.  If  we  attempt  to 
express  it  as  a  decimal,  the  figures  will  go  on  without  end. 

II.  The  ratio  ot  A  to  B  may  also  be  conceived  of  as  a 
number  expressing  the  magnitude  of  A  when  we  take  B  as  the 
unit  of  measure.  This  amounts  to  the  same  thing  as  I.,  be- 
cause when  we  multiply  unity  by  any  factor  we  produce  the 
factor  itself. 

III.  If  A  and  B  are  numbers,  instead  of  geometric  magni- 

tudes,  the  ratio  of  ^  to  5  is  the  quotient  -g. 

The  consistency  of  these  ways  of  conceiving  a  ratio  is  es- 
tablished by  the  following  definition: 

353.  JDef.  To  multiply  a  magnitude  5  by  a  nu- 
merical factor  r  means  to  find  a  magnitude  wMcli  shall 
have  the  same  rati  o  to  B  that  r  has  to  unity. 

Hence  the  expressions 

A  :  B  ::  r  :  1 

and 

A=rB 

are  equivalent. 

The  preceding  definition  of  a  ratio  gives  us  another  defini- 
tion of  .«  proportion,  namely: 

3  W.  Four  magnitudes  are  proportional  when  the 
ratio  of  the  §rst  to  the  second  is  equal  to  the  ratio  of 
the  third  to  the  fourth. 

355.  Bef.  If  the  terms  of  a  ratio  are  interchanged, 
the  new  ratio  is  called  the  inverse  of  the  original  one. 


|-i.~»t  i\ 


RATIOS  OF  INCOMMENSURABLE  MAGNITUDES.     169 


Thus  the  ratio  B  :  A\b  the  inverse  oi  A  :  B, 

It  A  :  B  ::  m  :  71,  then  A  :  B  =  ~   and  B  :  A  =  -:the 

mn  ^^  ^ 

product  of  these  ratios  is  —  =  1.    Therefore: 

nm 

356.  Theorem.  The  product  of  Uoo  inverse 
ratios  is  unity. 

Ratios  of  Incommensurable  Magnitudes. 

357.  If  two  magnitudes  are  incommensurable  (§  336), 
they  may  still  be  considered  as  having  a  ratio,  but  this  ratio 
cannot  be  exactly  expressed  by  a  fraction.  Let  us  suppose 
that  in  dividing  the  magnitude  B  into  n  parts,  A  is  found  to 
contain  m  of  these  parts  and  a  fraction  of  another  part. 

Then  the  ratio  of  ^  to  J5  will  be  greater  than  -,  and  less 

Wl  -f-  1  W2  1  ^ 

than ;  that  is,  less  than  — \-  -.    The  number  n  may 

n  n       n  *' 

here  be  as  great  as  we  please. 

358.  Theorem.  If  four  incommensur  able  magni- 
tudes A,  B,  P,  and  Q  are  so  related  that,  on  dividing 
the  antecedents  A  and  P  each  into  n  equal  parts,  Q 
shall  contain  the  same  whole  number  of  parts  if  P 
that  B  contains  of  A^  fractions  being  neglected,  and 
this  however  great  the  number  n,—then  the  ratio  of 
Qto  P  is  equal  to  the  ratio  of  B  to  A^  and  A,  B^  P, 
and  Qform  a  proportion. 

Hypothesis,        ^  B  <A  <  (- +1]  B, 
n  \n      n) 


-Q<P 


f+a«' 


how  great  soever  the  numbers  m  and  n. 

Conclusion.     A  :  B  ::  P  :  Q. 

Proof.     If  the  ratios  ^  :  J5  and  P  :  ^  be  unequal,  let  a 
be  their  difference.     Since  we  can  make  the  number  n  sib 

gro-.t  as  ,fo  please,  let  us  make  it  so  great  that  -  shall  I;>e  lees 
than  a  (§  345,  Ax.  1). 


Ill 


11 


mmi ' 


M  ! 

\r4 


m 


■'ii  I 


HJIiii 


II  -ili 


!lll|! 


170        BOOK  V.     PROPORTION  OP  MA0NITUDB8. 

If  m  be  the  whole  number  of  times  which  A  contains  the 
»th  part  of  B, 

A:B>-  andA  :  B  <- 4--. 
n  n      n 

By  hypothesis  P  contains  the  wth  part  of  Q  this  same 
number  m  of  times,  plus  a  fraction.     Therefore 

P  '  Q>  -  and  P  :  Q  <  — h  - . 
n  n      n 

Since  both  ratios  are  greater  than  ^-  and  less  than  -4--, 

n  n  ^  )i' 

their  difference  ra:ijt  be  lesb  than  -  and  therefore  less  than 

n 

a,  because  ~  <ot.  Therefore  the  ililTerence  of  the  ratios  would 

be  at  the  same  time  equal  to  a  and  less  than  a,  which  is 
absurd.     Therefore  the  ratios  do  not  differ  at  all. 

359.  Corollary.  If  any  theorem  respecting  the  equality 
of  ratios  be  proved  for  the  ratios  of  all  comme?isurable  magni- 
tudes, however  small  the  common  measure,  it  will  hold  true 
for  the  ratios  of  all  incommensurable  magnitudes. 

360.  Bef.  The  ratio  of  two  incommensurable 
magnitudes  is  called  an  irrational  number. 

Although  an  irrational  number  cannot  be  expressed  as  the 
quotient  of  two  entire  numbers,  yet  by  taking  such  numbers 
sufficiently  great  we  can  find  quotients  which  shall  come  as 
near  as  we  please  to  the  irrational  number.     Thus; 

1 


To  come  within 
To  come  within 


1000 
1 


we  take  a  divisor  >  1000. 


-  we  take  a  divisor  >  100000, 


100000 

etc.  etc.  etc. 


Transformation  of  Proportions. 

361.  Def.  Inversion  is  when  the  terms  of  each 
ratio  in  a  proportion  are  interchanged  to  form  a  new 
proportion. 


TRANSFORMATION  OF  PROPORTIONS. 


Ill 


^1 

B  : 


Theorem  of  Inversion.    From  the  proportion 

A  :  B  .'.  P  :Q 
we  may  conclude  by  inversion 

B  '.Av.Q'.P, 
Proof.    From  §  355. 

363.  Def.  Alternation  is  when  the  means  of  a 
proportion  are  interchanged  to  form  a  new  proportion. 
The  new  proportion  is  then  said  to  be  the  alternate 
of  the  original  proportion. 

Theorem  op  Alternation".  In  any  proportion  the 
antecedents  Moe  the  same  ratio  to  each  other  as  the 
consequents. 
Hypothesis.     If 
A  :  B  ::  P  :  Q— 

Conclusion.    Then       ^  j ^ ; 

A:  P  ::  B:  Q.         q 

Proof.    1.       Let  the      '  '  ' 

ratios  A  :  B  and  P  :  Q  each  be  -,  so  that 

mth.  part  of  ^  =  nth  part  of  B,  which  call  a. 
mth.  part  of  P  =  ^ith  part  of  Q,  which  callj?. 

2.  Then  A  =  ma.        B  =  na. 

P  =  mp.        Q  =  npo 

3.  Hence  A  :  P  ::  ma  :  mp  ::  a  : p;  ) 

B  :  Q  ::  na  :  np  ::  a  :p.  ) 

4.  Therefore  A  :  P  ::  B  :  Q.     Q.KD. 

363.  Qorollary.  If  the  extremes  he  interchanged,  the  pro- 
portion will  still  be  true. 
Fcr  by  alternation  we  have 

A  '.  P  '.:  B  '.Q, 
and  then  by  inversion,  and  putting  the  second  ratio  first, 

Q:B::P:A. 


(§346) 


fj; 


\l 


I 


' 


! 


I 


364    B^f 

antecedents  and  consequents  are  compared  with  either 
antecedents  or  consequents  to  form  a  new  proportion. 


^_LJ 

L 

I 

;| 

1 

■i' 

m 

T 


mil 


I II 


i 


172         BOOK  V.     PROPORTION  OF  MAGNITUDES. 

Theorem  op  Composition.    If  we  have  the  proportion 

A  :  B  ::  F  :  Q,  (1) 

we  may  conclude 


A  :  A  +  B  ::  P  :  P+Q;) 
A-{-B:  B::  P^Q:  Q.\ 


(2) 


Proof.  Let  the  equal  ratios  in  (1)  be  m  :  n.  Using  the 
same  notation  as  in  the  preceding  theorem,  we  find 

A=ma\  B  =  na.      A-\-  B  =  {m-{-  n)a, 
P  =  mp',  Q  =np.      p  -f.  Q  =  (m  _|_  n)p, 

A  :  A-\-  B  =  m  '.  m-\-n. 

P  :  P-\-  Q  =  m  :  m-^-n 

A-\-  B  :  B  =  m  -\-  n  :  n, 

P  -{■  Q  '  Q  =  m  -{-  n  :  n. 

Whence  the  conclusions  (3)  follow  by  comparing  the  equal 
ratios. 

365.  Bef.  Division  is  when  the  difference  of 
antecedents  and  consequents  is  compared  with  either 
antecedents  or  consequents  to  form  a  new  proportion. 

Theorem  of  Division.    If  we  have  the  proportion 

A:B::P:Q,  (1) 

we  may  conclude 

A:A-B::P:P-Q;l  Y9\ 

A-B  :B  ::P- Q  :Q.\  ^'^> 

Proof.  By  the  same  process  as  in  the  last  theorem. 

366.  Theorem.    If  we  have  the  several  proportions 

A    :  B  ::  P    ;  Q, 
A'  :  B  ::  P'  :  Q, 

etc.  etc., 
we  may  conclude 

A  +  A'-i-  etc.  :  5  ::  P  +  P'  4-  etc.  :  Q. 

Proof.    The  proportions  show  that  if 


A 


n 


m 


A*  =  —iB.  etc. 


n' 


MULTIPLE  PHOPOBTIONS. 


173 


then 


whence 


P'  =  %Q,  etc., 


^  +  ^'  +  etc.  =  (^  +  ^'  +  etc.)i?, 

P  +  P'  +  el..  =  (^+-'  +  ctc.)c. 

The  conclusion  now  follows  from  the  equality  of  the  co- 
efficients. 

Multiple  Proportions. 

367.  When  three  or  more  ratios  are  equal,  a  proportion 
may  be  formed  between  any  two  of  them.     Thus,  if 

A:B  =  M:N=P'.Q=.XiY,  etc.,  (1) 

we  may  form  the  proportions 

A:B::M:W, 
A'.B'.'.XiY, 
M:N'.:X'.Y, 

etc.        etc. 
The  equality  of  such  ratios  is  generally  expressed  by  writ- 
ing all  the  antecedents  with  the  sign   :   between  them,  fol- 
lowed by  the  consequents  in  the  same  order. 
Thus  (1)  would  le  expressed  in  the  form 

A  :M:P'.X=B:N:Q:  Y,)  ,., 

or  A'.M:P:X',:B:NiQ:Y.)  ^^ 

Here  the  first  consequent  {B)  corresponds  to  the  first 
antecedent  {A)-,  the  second  {N)  to  the  second  (if),  etc. 

368.  Bef.  A  proportion  of  six  or  more  terms 
expressed  in  tlie  form  (2)  is  called  a  multiple  propor- 
tion. 

Simple  proportions  may  be  formed  as  follows  from  a 
multiple  proportion. 

369.  Theoeem.  Ill  a  multiple  proportion  any 
antecedent  is  to  its  consequent  as  any  other  ante- 
cedent to  its  consequent. 


mm 


i     '■ 

i ' 
1 

174         BOOK  V.     PliOPOHTiijN  OF  MAGNITUDES 

Example.  In  the  proportion  (2),  us  the  first  antecedent,  A, 
is  to  the  first  consequent,  B,  so  is  the  third  antecedent,  P>  to 
the  third  consequent,  Q\  or,  in  symbolic  language, 

A'.BwP'.Q, 

Proof,  This  theorem  follows  at  once  from  the  form  of 
expression  in  (1)  and  (2). 

370.  Theorem.  In  a  multiple  proportion  any 
two  antecedents  are  to  each  other  as  the  correspond- 
ing consequents. 

Example.  In  (2),  as  A  is  to  P,  so  is  B  (the  consequent  of 
^)  to  ^  (the  consequent  of  P);  or 

A'.P'.'.B'.Q. 

Proof.  Any  such  proportion  is  the  alternate  of  one  of  the 
original  proportions  expressed  by  the  continued  proportion. 
Thus  one  of  the  original  proportions  expressed  in  (1)  is 
A  :  B  ::  P  :  Q,  and  the  above  proportion  is  its  alternate. 

371.  Theorem.  In  any  proportion  the  sum  of 
any  number  of  antecedents  is  to  the  sum  of  the 
corresponding  consequents  as  any  one  antecedent  is 
to  its  consequent. 

Proof.  Let  the  proportion  be  that  in  (2),  and  let  the  ratio 

of  each  antecedent  to  its  consequent  be  — ,  so  that 

n 

A  :  B  ::  m  :  n, 

M  '.  N ::  m  :n, 

P  :  Q  ::  m  :n, 

etc.       etc. 


Then 


B 


A 
m 

m       n 


N 


m 


n 


etc.  etc. 


MULTIPLE  PROPORTIONS. 


17C 


By  adding  these  equations  wo  have 

A^M-^r  -\- etc.  __  yy -f  jyr 4- g 4- etc. 
m  n  > 

that  is,  the  wth  part  of  ^  +  Jf  +  P  +  etc.  is  oqu.ii  to  the 
wth  part  oiB  -\-  N  -\-  Q  -{-  etc.,  so  that 

A  +  Jf  +  P  +  etc.  :  /y  +  jv+  g  +  etc.  ::  m  :  n, 
which  is  by  hypothesis  t..u   ^ume  as  the  ratio  of  each  ante- 
cedent to  its  consequent. 

Because  this  reasoning  is  correct  how  great  soever  wo  sup- 
pose the  numbers  m  and  7i,  the  theorem  is  true  whether  the 
magnitudes  are  commensurable  or  incommensurable  (§  359). 

372.  Theorem.  In  any  proportion  the  difference 
of  any  two  antecedents  is  to  the  difference  of  the  cor- 
responding consequents  as  any  antecedent  is  to  its 
consequent. 

Proof.  By  taking  the  difference  of  the  first  two  equations 
of  §371,  we  have 

A  - M _ B- N 

m      ~      n     ' 
which  shows  that 

A  —  M  :  B  —  ]^ ::  7n  :  n, 

the  same  ratio  which  each  antecedent  has,  by  hypothesis,  to 
its  consequent. 

In  the  same  way  it  may  be  shown  that  any  other  difference 
of  the  corresponding  magnitude  has  this  ratio. 

373.  Theorem.  If  in  a  series  of  ratios  the  conse- 
quent of  each  is  the  antecedent  of  the  next,  the  ratio 
of  the  first  antecedent  to  the  last  consequent  is  equal 
to  the  product  of  the  separate  ratios. 

Hypothesis.     We  have  the  separate  ratios 

A:B. 
B  :  C. 
GiD. 

Oonchtsion.  The  ratio  of  A  to  D  is  the  product  of  the 
ratios  A  :  B,  B  :  C,  C  :  D. 


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176  BOOK  V.     PROPORTION  OF  MAGNITUDES. 

Proof,    Let  the  values  of  the  respective  ratios  A  :  B, 

B:G,0:Dhe-,  i,  A  so  that 
w    J    q 

A:  B  ::7n:n,0T  ratio  A  :  B  =  ^, 

n 

B  :  0::  p:q,0T  ratio  B  :  C=^, 


Then 


G  iB  ::i  :  ;,  or  ratio  C  :  i>  =  4-. 

J 


A 
m 
B 
P 

i 


B  ^ 

q' 

•  • 

.7 


(1) 


Divide  the  first  equation  by  p  and  the  second  by  n, 

mp~'  np*  np  ^  nq' 

A__G_ 
mp~  nq' 
Divide  this  equation  by  i  and  the  last  of  (1)  by  nq, 


Therefore 


G 


Therefore 
that  is. 


mpt       nqi 


A 


nqi 
D 


D 

nqj 


mpt       nqi ' 
nqj        n       q  ^  i 


374.  Def.  When  the  ratio  of  the  first  antecedent 
to  the  last  consequent  is  formed  by  multiplying  a 
series  of  intermediate  ratios,  the  ratio  thus  obtained  is 
said  to  be  compounded  of  these  intermediate  ratios. 


LINEAM  PROPORTIONS, 


111 


CHAPTER    II. 

LINEAR    PROPORTIONS. 


Definitions. 

375.  Def.    Similar  figures  are  those  of  which  the 
angles  taken  in  the  same  order  ^ 
are  equal,  and  of  which    the 
sides  between  the  equal  angles 
are  proportional. 

Example.    The  figure  A  BCD 
is  similar  to  A'B'C'D'  when 
Angle  A  =  angle  A'y 
Angle  B  =  angle  B%  etc., 
and 

AB  :  BO  :  CD  :  DA  ::  A'B' :  B'C  :  CD'  :  D'A\ 

376.  Def.  In  two  or  more  similar  figures  any- 
side  of  the  one  is  said  to  be  homologous  to  the  corre- 
sponding side  of  the  other. 

Example.    In  the  above  figr  re, 

the  sides  AB  and  A'B'  are  homologous, 

the  sides  BG  and  B'O'  are  homologous, 

etc.  etc.  etc. 

377.  Bef.  When  a  finite  straight  Kne,  as  AB,  is 
cut  at  a  point  P  be- 

tween  A  and  B  it  is  ! f       i  ? 

said  to  be   divided^ 


B 


intenially  at  P,  and  the  two  parts  AB  and  BB  are 
called  segments. 

378.  Bef.    If  the  straight  line  AB  is  produced 
and  cut  at  a  point  Q  outside  of  A  and  P,  it  is  said  to 
06  divided  eztemallv  at  a  an^  f>»^  Hr. 
-o  V  are  called  segments. 


mU 


178 


BOOK  V.     PROPORTION  OF  MAGNITUDES. 


Corollary  1.  A  line  cut  internally  is  equal  to  the  sum  of 
its  segments. 

Cor.  2.  A  line  cut  externally  is  equal  to  the  difference  of 
its  segments. 

379.  Bef.    Two  straight  lines    are    said    to   be 
similarly  divided  when  the  diiferent  segments  of  the 
one  have  the  same  ratios  as         c  n        d 
the  corresponding  segments                "  ~    "• 
of  the  other.                             -J M  b 

Example  1.  If  the  line  AB  m  divided  at  M  and  the  line 
cd  at  N  in  such  manner  that 

AM:  MB  ::  CN :  ND, 
the  lines  A  B  and  CD  are  similar- 
ly divided  at  M  and  N.  C  N         Q  ly 

Example  3.  If  the  lines  AB 

and  CD  are  divided  at  M,  P,  N,  A M Y    3 

and  Q  in  such  wise  that 

AM :  MP  :  PB  ::  CN :  NQ  :  QD, 
they  are  similarly  divided. 

380.  Def.  If  three  straight  lines,  a,  5,  c,  are  so 
related  that 

a  \h  '.',h  :  c, 
the  line  h  is  said  to  be  a  r^'^an  pro- 
portional between  a  and  c. 


01- 


Theorem  I. 

381.  If  two  straight  lines  are  similarly  divided^ 
each  part  of  the  first  has  the  same  ratio  to  the  corre- 
sponding part  of  the  second  that  the  whole  of  the  first 
has  to  the  whole  of  the  second. 

Hypothesis.  Two  straight  ^ 

lines,  AB  and  A'B\  divided  '~ 

at   P,    Q,   P',   and  Q',   so  A^ p'  tf    g 

that 


i. 


0  B 

I  ■  ■  I 


AP  :  PQ'.  QB  ::  A'P'  :  P'Q'  :  Q'B\ 
Conclusion.    AP  :  A'P'  ::  AB  :  A'B',  \ 
PQ:  P'q  ::  AB  :  A'B'-A 


LINEAR  PRQP0UTI0M8. 


179 


0,  are  so 


or,  expressed  as  a  multiple  proportion, 

AF  :  FQ:  QB  :  AB  ::  A'F'  :  F'Q'  :  Q'B'  :  A'B\ 
Froof.     In  the  proportion  of  the  hypothesis  the  sum  of 

the  antecedents  is  AB,  and  the  sum  of  the  consequents  A'B' 

Therefore  (§371) 

AB  :  A'B'  ::  AP  :  A'F'  ::  FQ  :  F'Q\  etc.     Q.E.D. 

Theoeem  II. 

383.  A  line  cannot  be  divided  at  two  different 
points,  both  internal  or  both  external,  into  segments 
having  the  same  ratio  to  each  other. 

Hypothesis  I.     A  line,  AB, 

divided  internally  at  the  points  ^ 

P  and  Q, 

Conclusion.  The  ratio  AF  :  FB  will  be  different  from 
the  ratio  A  Q  :  QB. 

Fro^f.  Let  the  ratio  AF  :  FB  be  ^.  Then  JP  will  con- 
tain m  parts,  and  FB  n  equal  parts. 

Because  AQi^  greater  than  AF,  it  will  contain  more  than 
m  parts;  and  because  QB  is  less  than  FB,  it  will  contain  less 
than  n  parts. 

Therefore  the  numerator  of  the  ratio  AQ  .QB  will  be 
greater  than  w,  and  its  denominator  less  than  n,  whence  it 

must  be  greater  than  -  and  cannot  be  equal  to  it. 

Therefore  there  is  no  other  point  of  division  than  F  for 
which  the  ratio  of  the  segments  will  be  the  same  as  ^P  :  FB. 

Hypothesis  II.  A  line,  AB,  divided  externally  at  the 
pomts  F  and  Q. 

Conclusion.        The     ratio    ^^- ^  ^     ^ 

AF  :  BF  will  be  different  from  the  ratio  AQ  :  BQ. 

Froof.  Let  m  be  the  number  of  equal  parts  in  AF:  n,  the 
number  in  BF-,  and  s,  the  number  in  FQ.    Then 

m 


AF  :BF  = 


^Q 


n 
n~{-8 


180 


BOOK  V.     PROPORTION  OB'  MAGNITUDES. 


If  we  reduce  these  fractions  to  a  common  denominator  and 
take  their  difference,  we  find  it  to  be 

{m  —  n)8 
n  {n-\-s)* 
Because  m  and  n  are  necessarily  different,  this  fraction 
cannot  be  zero,  and  the  ratio  AP  :  BP  is  different  from 
AQiBQ.    Q.E.D. 

383.  Corollary  1.  When  the  point  of  division  P  is  nearer 
to  B  than  to  A,  AP  is  greater  than  BP,  and  the  ratio 
AP  :  BP  is  greater  than  unity. 

When  it  is  nearer  to  A  than  to  B,  AP  is  less  than  BP, 
the  ratio  is  less  than  unity. 

384.  Cor.  2.  If  we  suppose  the  point  P  to  move  from 
A  toward  B,  the  ratio  AP  :  PB  will  be  equal  to  zero  as  P 
starts  from  Al  will  be  unity  when  P  is  half  wa}  between  A 
and  B,  and  will  increase  without  limit  as  P  approaches  B. 

385.  Cor.  3.  A  line  cannot  be  divided  externally  into 
segments  having  the  ratio  unity. 

386.  Cor.  4.  Two  different  points  may  be  found,  the 
one  internal  and  the  other  external,  which  shall  divide  a  line 
into  segments  having  the  same  given  ratio. 

EXERCISES. 

1.  Draw  a  line,  AB,  and  cut  it  internally  in  several  points 
so  that  the  ratios  of  the  segments  shall  be 

1  :  6,  2  :  5,  3  :  4,  4  :  3,  5  :  2,  6  : 1. 

2.  Cut  the  same  line  externally  in  the  ratios 

2  :  9,  1  :  8,  8  :  1,  9  :  2,  11  :  4. 

3.  A  line  7  inches  long  is  to  be  divided  into  segments  hav- 
ing the  ratio  4  :  5.     How  lo'ng  are  the  segments? 

4.  A  line  6  inches  long  is  to  be  divided  externally  into 
segments  having  the  ratio  5  :  8.  How  far  is  the  point  of 
division  from  each  end  of  the  line. 

5.  If  the  line  ^5  (§  377)  is  3  centimetres  in  length,  and 
the  points  P  and  Q  divide  it  both  internally  and  externally 
into  segments  having  the  same  ratio  1 :  2  (§  386),  find  the 
lengths  AP,  AB,  and  AQ. 


LINEAR  PROPORTIONS.  131 

Theorem  III. 

387.  fftwo  straight  lines  are  cut  by  three  or  more 
parallel  straight  lines,  any  two  intercepts  on  the  one 
are  as  the  corresponding  intercepts  on  the  other. 

Hypothesis,  a,  by  c,  three  parallels  intersecting  the  line 
p  in  the  points  A,  B,  G,  and  the  « 

line  q  in  A\  B',  G\  y      V 

Conclusion.  «■  "^^ 

AB  :  BG  ::  A'B'  :  B'C\ 

Proof.    Diyide  AB  into  any  ^  b/" — 1l^ 

number  m  of  equal  parts,  and 

through  the  points  of  separation 

draw  lines  parallel  to  AA^  and  *:  /n  /r' 

BB\  /  i 

Cut  off  from  BC  parts  equal  to  those  of  AB.  Let  the 
number  of  parts  in  BC  be  n  plus  a  fraction.  Through  the 
pomts  of  separation  draw  lines  parallel  to  BB\    Then— 

Because  the  lines;?  and  q  are  cut  by  parallels  intercepting 
equal  lengths  on  p,  the  intercepts  on  q  are  also  equal  (§132, 1. ). 

Therefore  A'B'  is  divided  into  m  equal  parts,  and  B'C  is 
divided  into  n  equal  parts  plus  a  fraction. 

Because  this  is  true  however  great  the  numbers  m  and  n, 
we  conclude 

AB  :  BC ::  A'B'  :  B'C  (§  358).     Q.E.D. 
Corollary.    If  the  points  A  and  A'  coincide  so  that  »  and 
q  cross  each  other  at  A,  the  figure  4 

AGC  will  form  a  triangle;  and 
the  conclusion  will,  by  the  same 

demonstration,  be  b/ X^b> 

AB  :  BG  ::  AB'  :  B'G*.         ^^_ >^q, 

But55'  is  parallel  totheside  CC'ot  the  triangle.  Therefore: 

388.  A  straight  line  parallel  to  one  side  of  a  triangle 
divides  the  other  two  sides  similarly. 

Theoeem  IV. 

^-  ^^^'  I(  ^^^  **^^*  ^-^  ^  i'Tiangle  he  similarly 
aimaed,the  line  joining  the  points  of  division  will 
uoparauec  to  the  rmaining  sids  of  the  triangle. 


i  Hi 


189 


BOOK  V.     PROPORTIOJSr  OF  MAQNITUDEa 


Hypothesis.    ABCy  a  triangle  of  which   the  sides  CA 
and  CB  are  cut  (internally  or 
externally)  in  the  points  D 
and  E  in  such  manner  that 

CD  :  AD  ::  CE :  BE. 


Conclusion.  DE\\AB, 

Proof.      If   DE  is    not 
parallel  to  AB,  draw  DE' ^ 
parallel  to  ^^  and  meeting 
CB  in  E".    Then 

CD:  AD::  CE'  :  BE',  (§  388) 

and  the  line  CB  is  divided  at  two  points,  E  and  E'l  into  parts 

haying  the  same  ratio  CD  :  AD,  which  is  impossible  (§  382). 

Therefore  the  points  E  and  E'  coincide,  and  the  line  DE 

is  the  same  as  the  parallel  i>ii^'.    Q.E.D. 

Theokem  V. 

390.  Equiangular  triangles  are  similar,  and  the 
sides  between  the  equal 
angles  are  homologous 
to  each  other. 

Hypothesis.  ABC  and 
DEF,  equiangular  triangles 
such  that 

Angle  C  =  angle  F. 

Angle  A  —  angle  D. 

Angle  B  =  angle  E. 

Conclusion.        AB  : 
AB  : 
AB:  BC: 


or 


BC  ::  DE  :  EF, 
AC::  DE:  DF, 
CA  ::  DE:  EF  :  FD. 
Proof.    Apply  the  triangle  EDF  to  ABC  so  that  -P shall 
coincide  with  C  and  FD  shall  fall  on  CA,    Let  D'  be  the 
point  of  CA  on  which  D  falls.     Then — 
1.  Because  angle  F  =  angle  C, 

UneFE=CB. 
ijui;  Jii   ue  ine  poiux,  on  wniun  ^  laus. 


LINEAR  PROPOJEiTIONa.  jgg 

4.  Therefore  '''^"^^-  (8  69) 

CZ)'  :  AD'  ::  C^'  ;  BE'; 
and,  by  composition  (§  364), 

GD'  :  CA  ::  GE' :  GB; 
or,  because  GD'  =  FD  and  Cj^'  =  FE, 

T    .1,  I>F '.  AG ',x  EF '/bG,    Q.E.D. 

m  the  same  way  it  may  be  shown  that  the  other  nronnr 
tions  of  the  conclusion  are  true.  ^    ^'' 

391.  CW/ary.  ijT/rom  owe  ^n«w^/e  anotJier  he  cuf  nff 
by  a  hue  parallel  to  one  of  its  sides,  thetna^ttUuslut  S 
will  be  similar  to  the  original  triangle.  ^ 

Theoeem  VI. 

392.  If  the  sides  of  one  triangU  have  to  each 
other  the  same  ratios  as  the  sides  of  another   these 
triangles  are  equiangular  and  similar   '^''''^'  ^^''' 
Hypothesis.   Two  triangles,  ABG  and  DEF,  such  that 

AB  :  BG  :  GA  ::  BE  :  EF :  FD. 
Conclusions, 
Angle  6^  (opposite  AB)  =  angle  i^  (opposite  DE), 
Angle  A  {      «       BG)  =  angle  D(      <^        fF) 
Angle  B(      «        <7^)  =  angle  ^(      «        fD) 
Proof.   On  the  sides  FD  and  FEot  the  triangle  />;?;?» 
t^ke  the  points  M  and  N  such  that  ^        ^^ 

FM=:GA, 
EJ^  =  GB, 
and  join  MJV.    Then— 

1.  Because  FM  =  GA 
and  JYF  =  BG, 

and  because 

SGiGA  y.EF.FD  (hyp.), 
we  have  - 

ly'F :  FM  II  EF :  FD.  ^L 


Hi 


ll  lill 


!l!i 


184 


BOOK  V.     rUOPORTION  OF  MAQNITUDE8, 


%.  Therefore  the  sides  FD  and  FE  are  divided  similarly 
at  M  and  N,  whence 

MN  II  DE,  ^§389) 

3.  Because  these  lines  are  parallel,  the  triangles  FDE  and 
FMN  are  similar  (§  391).     Therefore 

MN  :  NF'.  FM  ::  DE  :  EF  :  FD,  (§  390) 

4.  Because  NF=  ^Cand  FM  -  GA, 

MN  :BC  :CA  ::  DE  :  EF  :  FD, 

6.  Comparing  with  the  hypothesis, 

MN  =  AB. 

Therefore  the  triangle  FMN  is  identically  equal  to  CAB^ 
and,  comparing  the  angles  opposite  the  equal  sides, 

Angle  G  =  angle  F. 

Ang\e  A  =  angle  FMN  =  angle  D; )  /„v 

7.  )  ^  ^ 

Q.B.D. 


Angle  B  =  angle  FMN  =  angle  ^. 


Theorem  VII. 

393.  Two  triangles  having  one  angle  of  the  one 
equal  to  one  angle 
of  the  other^  and 
the  sides  containing 
these  angles  propor- 
tional^ are  similar. 

Hypothesis.  ABG 
and  A'B'G'y  two  tri- 
angles in  which 

Angle  G  —  angle  C". 
aA'  :  G'B'  ::  GA  :  GB. 

Conclusion.    The  triangles  are  similar;  that  is. 
Angle  A  —  angle  A\ 
Angle  B  =  angle  B', 
AB'.BG:  GA  ::  A'B'  :  B'G'  :  C'A', 
Proof.  In  GA  take  GP  =  G'A'^  and  draw  PQ  parallel  to 
AB,     Then— 


LINEAR  PROPORTIONS. 


185 


I.  The  triangle  CAB  is  similar  to  CPQ.  (9,  391  \ 

11.   CPiOQy.CA  :GB.  (ggsg) 

2.  By  hypothesis,  CM'  :  O'B'   ::  CA  :  CB,  and  CP  = 
G'A'y  by  construction.     Therefore 

GP  :  G'B'  ::  GA  :  GB. 

3.  Comparing  with  (i),  II., 

G'B'  =  6'(>. 
Therefore  the  triangles  G'A'B'  and  CPg  have  two  sides 
and  the  included  angle  equal,  and  are  identically  equal. 

4.  Comparing«with  (1),  I., 

Triangle  G'A'B'  similar  to  triangle  GAB.    Q.E.D. 


Theorem  VIII. 

394.  Mectilineal  figures  similar  to  the  same  figure 
are  similar  to  each  other. 

Hypothesis.       Two      ^. 
figures,  P  and  Q,  each  r^\ 
similar  to  the  figure  A.\  \ 

Gonclusion.     P  and  i     ^      / 
Q  are  similar  to  each  \^       / 
other.  \    / 

Proof.     1.  Let  any         ^ 
side  of  P,  a'  for  in- 
stance, be   to  the   ho- 
mologous side  aotA  asmm,  and  the  side  a  be  to  the 
homologous  side  a'  of  ^  as  jt>  :  ^,  so  that 


a' 


m 
«  =  -, 
n 


a:  a''  =  ^. 


Then,  because  the  antecedent  of  one  of  these  ratios  is  the 
consequent  of  the  next, 


«-:a"  =  ^. 


nq  (§  373) 

In  the  same  way  it  may  be  shown  that  every  side  of  P  has  to 

tiie  nomoloorniia  airlQ  nf  /y  +>.«  ««,», j.z-   ... 

o ^''^  ^^  a  "iv  o»iuc  luiiu  mp  :  no. 


J  u 


\\ 


H' ■' 


186         SOOK  V.     PROPORTION  OF  MAQNITUDES. 

2.  Because  the  angles  of  P  and  Q  are  severally  equal  to 
the  corresponding  angles  of  A,  they  are  equal  to  each  other. 

3.  The  figures  P  and  Q  having  their  homologous  sides  in 
the  same  ratio  mp  :  nqj  and  their  angles  equal,  are  similar 
by  definition  (§  376).    Q.E.D. 

Theorem  IX. 

395.  Similar  polygons  may  he  divided  into  the 
same  nv/mber  of  similar  triangles. 

Hypothesis.  Two  simi- 
lar polygons,  ABCDE  and 
A'B'C'D'E',  divided  into 
triangles  by  lines  drawn 
from  the  corresponding 
angles  G  and  \C  to  all  the 
non-adjacent  angles. 
Conclusion. 

Triangle  CDE  similar  to  CD'E', 
Triangle  CEA  similar  to  G'E'A', 
etc.  etc. 
Proof.    1.   In  the   triangles    CDE  and  CD'E',  angle 
D  —  D*  (hyp.  and  def.),  and 

CD  :  C'Z)'  ::  DE  :  D'E\ 
Therefore  these  triangles  are  similar  and  equiangular,  and 

CE  :  G'E'  ::  CD:  G'D'  ::  DE  :  D'E*.      (§  393) 

2.  From  the  equal  angles  DEA  and  D'E'A'  take  the  equal 
angles  DEC  and  D'E'G',  and  we  have  left 

Angle  CEA  =  angle  G'E'A'. 

Also,  from  (1)  and  the  hypothesis  of  similarity  of  the  two 

figures, 

CE  :  CE'  ::  EA  :  E'A\ 

Therefore  the  triangles  CAE  smd  G*A'E*  are  also  similar. 

3.  In  the  same  way  it  may  be  shown  that  all  the  other 
triangles  into  which  the  polygons  are  divided  are  similar. 

Q.E.D. 

396.  Def.  Similar  figures  are  said  to  be  similar- 
ly placed  when  so  placed  that  each  side  of  the  one 
shall  be  T>arallel  to  the  homologous  side  of  the  other. 


LINEAR  PROPORTIONS, 


187 


Theorem  X. 

t 

307.  fftwo  similar  figures  are  similarly  placed, 
then — 

I.  All  straight  lines  joining  a  vertex  of  one  to 
the  corresponding  vertex  of  the  other  meet  in  a  point 
when  produced, 

II.  The  point  of  meeting  divides  the  lines  exter- 
nally into  segments  having  the  same  ratio  as  the 
ho7nologous  sides  of  the  figures. 


■••^> 


Hypothesis.     ABCD  and  A'B'C'D\  two  eimilar  figures 
in  which 

AB  '.  BC :  CD  :  DA  ::  A'B'  :  B'C  :  CD'  :  D'A% 

and  of  which  each  side  is  parallel  to  its  homologous  side  in 
the  other. 

Conclusion.    I.  The  lines  A  A',  BB',  etc.  (which  we  shall 
call  junction  lines),  being  produced,  meet  in  a  point  P. 
II.  AP  :  A'P  ::  BP  :  B'P,  etc.  ::  AB  :  A'B\ 
Proof.     1.  If  the  junction  lines  AA'  and  BB'  are  not 
parallel,  they  must  meet  in  some  one  point.     Let  P  be  this 
point. 

2.  Because  AB  and  A'B'  are  parallel,  the  triangles  PAB 
and  PA'B'  are  similar,  so  that 

AP  :  A'P  ::  BP  :  B'P  ::  AB  :  A'B\       (§391) 

3.  It  may  be  shown  in  the  same  way  that  if  we  call  0  the 


U' 


188 


BOOK  V.     PROPORTION  OF  MAGNITUDEQ. 


m 


1 1 


ii 


A'B'  (hyp.), 
AB  :  A'B\ 


point  in  which  the  junction  lines  BB'  and  CO'  meet,  we 

shall  have 

BQx  B'Q'.i  BG:  B'C'x  ^ 

or,  because  BG  :  B^G'  ::  AB 

BQ  :  B'Q  : 

4.  Comparing  with  (2), 

BQ  :  B'Q  ::  BP  :  B'P, 

C\  Therefore  the  line  BB'  is  cut  externally  at  P  and  at 
Q  into  segments  having  the  same  ratio;  namely,  the  ratio  of 
the  homologous  sides  of  the  figures. 

But  a  line  can  be  cut  only  at  a  single  point  into  segments 
having  a  given  ratio  (§  382).  Therefore  the  points  P  and  Q 
are  the  same;  that  is,  the  lines  A  A'  and  C(7'  cross  BB'  at  the 
same  point. 

6.  In  itie  same  way  it  is  shown  that  all  the  other  junction 
lines  intersect  dt  the  point  P,  and  that  the  segments  termi- 
nating at  P  have  the  same  ratio  as  the  homologous  sides  of 
the  figures.    Q.E.D. 

398.  Corollary.  If  the  similar  figures  are  equal,  the 
junction  lines  will  all  be  parallel  and  the  point  of  meeting 
will  not  exist. 

399^  Def.  The  point  in  wliich  the  lines  joining 
the  equal  angles  of  two  similar  and  similarly  placed 
figures  meet  each  other  is  called  the  centre  of  simili- 
tude of  the  two  figures. 

Theoeem  XI. 

400.  A  perpeniiczUar  from  the  right  angle  to  the 
hypothenuse  of  a  right-angled 
triangle  divides  it  into  two 
triangles,  each  similar  to  the 
whole  triangle. 

Hypothesis.  ABG.  a  triangle, 
right-angled  at  G;  GD,  a  perpen- 
dicular from  (7  on  AB. 

KyUfiVtaCSlUfi.       J-iiC     l/X iUiIlgilJB  JlJJX^,    -<Ul>X/,    UUU     \JJJU    iiiO  aii 

similar  to  each  other,  so  that 

GB'.GA'.  AB  ::.  DC  :  DA  :  AG  ::  DB  :  DG  :  BG. 


LINEAR  PB0P0BTJ0N8. 


189 


Proof  1.  In  the  triangles  ABC  and  A  CD  the  angle  A  is 
common,  and  the  angle  G  =  angle  D  because  each  of  them  is 
a  right  angle. 

Therefore  the  third  angles  are  also  equal,  and  the  tri- 
angles are  equiangular.  Comparing  the  sides  opposite  equal 
angles, 

CBiCAiAB  ::DC:DA:Aa    Q.E.D. 
3.  In  the  same  way  is  shown 

CBiCA:  AB  ::  DB -.  DC  :  BG.    Q.E.D. 
Corollary  1.    Comparing  the  equiangular  triangles  ADC 
and  GDB,  we  have 

AD  :  DC  ::  DC  :  DB. 

Therefore  DC  is  a  mean  proportional  between  AD  and 
DB,  or: 

401.  The  perpendicular  from  the  right  angle  upon  the 
hypothenuse  is  a  mean  proportional  between  the  segments  into 
tohich  it  divides  the  hypothenuse. 

Corollary  2.  It  has  been  shown  that 
lines  from  any  point  of  a  circle  to  the 
ends  of  a  diameter  form  a  right  angle 
with  each  other.     Therefore: 

408.  If  from  any  point  of  a  circle  a  perpendicular  he 
dropped  upon  a  diameter,  it  will  be  a  mean  proportional 
between  the  segments  of  the  diameter. 

Theorem  XII. 

403.  If  between  two  sides  of  a  triangle  a  parallel 
to  the  hase  he  drawn,   any  line 
from  the  'oertex  will  divide  the  base 
and  its  parallel  similarly. 

Hypothesis.  ABC,  &  triangle;  DE, 
a  parallel  to  AB,  intersecting  ^  C  in  D 
nnd  BG  in  E-,  GN,  a  line  from  G,  inter- 
bocting  DE  in  M  and  A  B  in  N. 

Conclusion.  DM  :  ME  .:  AN  i  NB. 

Proof.    1.  Because  in  the  triangles  GDM 


I 


n 


I 


190         BOOK  V.    PROPORTION  OF  MAGNITUDES. 

sides  i>if  and  AN  are  parallel,  these  two  triangles  are  equi- 
angular and  similar  (§  391). 

2.  Comparing  the  homologous  sides  opposite  the  equal 

angles, 

DM:  AN::  CM:  ON.  (§390) 

3.  In  the  same  way  it  may  be  shown  that  the  triangles 
CEM  and  CBN  are  similar,  so  that 

ME  :  NB  ::  CM  :  GN. 

4.  Comparing  with  (2), 

DM  :  AN  ::  ME  :  NBy 
or,  by  alternation, 

DM  :  ME  ::  AN  :  NB  (§  362).      Q.E.D. 

Corollary.  ltAN=  NB,  the  ratio  will  be  one  of  cqutJity 
and  we  shall  hajire  DM—  ME.    Therefore: 

404.  The  line  drawn  from  any  vertex  of  a  triangle  so  as 
to  bisect  the  opposite  side  will  also  bisect  any  line  in  the  tri- 
angle parallel  to  that  opposite  side. 

Theoeem  XIII. 

406.  The  bisector  of  an  interior  angle  of  a  tri- 
angle divides  the  opposite  side  into  segments  having 
the  same  ratio  as  the  two  adjacent  sides. 

Hypothesis.  ABC,  any  triangle; 
CDf  the  bisector  of  the  angle  C,  yt 

meeting  the  side  AB  in  D,  so  that 

Angle  AGD  =  angle  DGB. 

Conclusion. 

AD  :DB  ::AG:  CB. 

Proof.  Through  B  draw  BG 
parallel  to  DC,  meeting  AG  pro- 
duced in  G.    Then — 

1.  Because  DG  and.  BG  are  ■narallelj 

Angle  CGB  —  corresponding  angle  ACD. 
Angle  CBG  =  alternate  angle  DGB. 


.lO 


LtNEAR  PROPORTIONS. 


m 


3.  Comparing  with  the  hypothesis, 
Angle  OQB  .-  angle  CBG. 
Therefore  the  triangle  BQCis  isosceles,  and 

CB  =  CO, 
3.  Because  DC  and  BG  are  parallel, 
AD'.DB  V.AC'.  CG, 

:;  AC :  CB  (hy  2). 


Q.E.D. 


M 


Theobem  XIV. 

406.  The  bisector  qf  an  exterior  angle  of  a  tri- 
angle dimdes  the  opposite  side  exterimlly  into  seg- 
ments having  the  same  ratio  as  the  two  adjacmt 
sides. 

Hypothesis.    ABC,  9,  triangle;  CM,  the  continuation  of 
AG-y   CD,  the  bisector  of  the 
exterior  angle  at   C,  meeting 
AB  produced   in  D,  so  that 
angle  BCD  =  angle  MCD. 

Conclusion. 
AD.BD  ::AC:BC.        A- 

Prooof.  Through  B  draw 
BG  parallel  to  DC,  meeting  ^C  in  G.  The  proof  will  then 
be  so  much  like  that  of  Theorem  V.  that  it  is  left  as  an  exer» 
else  for  the  student. 

Corollary.  Since  the  bisectors  of  the  interior  and  ex- 
terior angles  of  a  triangle  ^ 
each  divide  the  opposite 
side  into  segments  having 
the  ratio  of  the  other 
two  sides,  the  ratios  of  the 
two  divisions  are  the  same. 
That  is.                                A- 


^' 


N 


P       B 
AP  :BP  iiAQ'.BQ. 


Q 


4.A7 


VV  horn     Q      lino   la    rlliri<1/irl    lT>+rt-t»*»on-rr    n-nA 

'»    ii'-^-J*     X.V     j-XXiT_-    i!j    ^J-i  T  X-vA^^XJ.    J.XXI/T7X  J^CtiJ_L jr      CXiXXVL 


externally  into  segments  having  the  same  ratio,  it  is 
said  to  be  divided  harmonically. 


192 


BOOK  V.     PBOPOBTION  OF  MAONITUDM. 


The  preceding  corollary  may  therefore  be  expressed  thus: 

408.  The  bisectors  of  an  interior  and  exterior  angle  at 
the  vertex  of  a  triangle  divide  the  base  harmonically » 

Theorem  XV. 

409.  If  a  line  AB  he  dimded  harmonically  at  the 
points  P  and  Q,  the  line  PQ  will  he  divided  harmom- 

cally  at  the  points  A  and  B,  x  P    b  Q 

t 1 — f H 

Hypothesis.  A  line  AB  divided  internally  at  P  and  ex- 
ternally at  Q,  so  that 

AP  :  BP  ::  AQ  :  BQ, 
Conclusion.       PB  :  QB  ::  PA  :  QA, 

Proof.  From  the  ratio  of  the  hypothesis  we  have,  by  in- 
version, 4 

BP  :  AP  ::  BQ  :  AQ. 
Then,  by  alternation, 

BP:  BQ  ::  AP  :  AQ.    Q.E.D. 

HarnK^nic  Points. 

410.  Def.  The  four  points  A,  B  and  P,  ft  of 
which  each  pair  divide  harmonically  the  line  termina- 
ted by  the  other  pair,  are  called  four  harmonic  points. 

Scholium.  The  relation  of  four  harmonic  points  may  be 
made  clear  to  the  beginner  by  supposing  the  line  terminating 
in  one  pair    to  partly  „ 

overlap  the  line  termi-   r- — ^         i       i 

nating   in    the    other;  ^  ^ 

thus,  when  the  points  are  harmonically  aranged,  the  line  AB 
is  divided  harmonically  at  the  points  P  and  Qy  and  the  line 
P^  at  the  points  A  and  B. 

Theorem  XVI. 

411.  The  hypothenuse  of  a  right-angled  triangle 

in  difiidp.d  harmn.nnio.nll'H  h'ii  n/n/ii  nnn.ir  nf  l/ivifis  tTivnoinh 

the  right  angle,  making  equal  angles  with  one  of  the 
sides. 


LINEAR  PROPORTIONS. 


193 


Hypothesis.    ABC,  a  triangle,  right-angled  at  Ci  CM, 
CN,  two  lines  from  G,  mak- 

ing  angle  MCA  =  ACN,  ^d 

and  meeting  the  base  in  M  .,-''' 

and  JV. 

Conclusion.  The  base 
AB  is  divided  harmoni- 
cally at  M  and  N. 

Proof.  Produce  MC  to  **'  A      ^ 

any  point  B.    Then— 

1.  Because  angle  MCA  =  ACN,  CA  is  the  bisector  of 
MCN.  Therefore  CB,  perpendicular  to  CA  by  construction, 
IS  the  bisector  of  the  exterior  angle  NCD  (§  83). 

3.  Because  CA  and  CB  are  the  bisectors  of  an  interior 
and  exterior  angle  of  the  triangle  MCN,  the  base  MN  of  this 
triangle  is  divided  harmonically  at  the  points  A  and  B  (§408), 

3.  Therefore  the  line  AB  \^  divided  harmonically  at  the 
points  M  and  JST  (§  409).     Q.  E.  D. 

Theorem  XVII. 

413.  The  diameter  of  a  circle  is  dimded  har- 
monically hy  any  tangent  and  a  perpendicular 
passing  through  the  point  oftangency. 

Hypothesis.     AB,  a  diameter  of  a  circle;  CT,  a  tangent 
at  T,  cutting  the  diameter  (pro- 
duced) in  C;  77),  a  perpendicu- 
lar from  Tupon  AB. 

Conclusion.     The   diameter 
AB  \^  cut  harmonically  at   C^* 
andZ). 

Proof.  Join  TA  and  TB, 
and  produce  TD  until  it  cuts 
the  circle  in  C7".     T5  ->  — 

1.  Because  AB  is  a  diameter  perpendicular  to  the  chord 
TU,  it  bisects  the  arc  TU  (§231).     Therefore 

Arc  TA=z\  arc  TU=  arc^  £/: 
3.  Also,        Angle  ClA  =  ^  arc  TA.  (§334) 

Angle  ^  77)  =:i  archer.  (§235) 

Therefore  Angle  CTA  =  angle  ^77). 


ill 

I  r 


194         BOOK  V.    PROPORTION  OF  MAGNITUDES. 

3.  Because  the  angle  ATB  is  inscribed  in  a  semicircle,  it 
is  a  right  angle.  And  because  the  angles  CTA  and  A  TD  on 
each  side  of  TA  are  equal,  the  line  AB  is  divided  harmoni- 
cally at  C  and  i)  (§  411).     Q.E.D. 


»»i 


CHAPTER    ML 

PROPORTION  OF  AREAS. 


B 


Theorem  XVIII. 

413.  If  two  rectangles  have  equal  altitudes,  their 
areas  are  to  each  other  as  their  bases. 

Hypothesis.  ABGD  and  PQB8,  two  rectangles  in  which 
the  altitude  PR  is  equal  to  the    c  D 

altitude  A  G. 

Conclusion, 
Area  P8 :  area  ADi.PQ:  AB, 

Proof.  1.  Let  PQ  and  AB 
be  to  each  other  asm  :  n.  Then, 
if  P^  be  divided  into  m  parts 
and  AB  into  n  parts,  the  mth 
part  ot  PQ  will  be  equal  to  the 
nth.  part  of  AB. 

Through  the  points  of  division  m  parte, 

draw  lines  parallel  to  the  sides  of  each  rectangle. 

Then  the  area  PS-  will  be  divided  into  m  equal  parts  and 
AD  into  n  parts,  each  equal  to  the  parts  of  PS,    Therefore 
Area  PS  :  area  AD  ::  ^Q  :  AB, 

2.  Because  this  proportion  is  true  how  great  soever  may 
be  the  numbers  m  and  n,  it  remains  true  when  the  sides  AB 
and  PQ  are  incommensurable.     Therefore  the  proportion 

Area  PS  :  area  AD  ::  PQ  :  AB 
is  true  in  all  cases  (§  359).     Q.E.D. 

Corollary  1.  Because  the  area  of  a  triangle  is  one  half  the 
area  of  a  rectangle  having  the  same  base  and  altitude  (§  301), 


• 

i 

n  parts. 

B 

s 

rl              I 

T 

t       ! 

L 

1 

Q 

AREAS, 


195 


and  because  aliquot  parts  of  magnitudes  have  the  same  ratio 
as  the  magnitudes  themselves  (§  347),  therefore: 

414.  The  areas  of  triangles  having  equal  altitudes  are  to 
each  other  as  their  bases. 

415.  Cor.  2.  The  areas  of  all  triangles  having  their  ver- 
tices in  the  same  point  and  their  bases  in  the  same  straight  line 
are  to  each  other  as  their  bases. 

For  all  such  triangles  have  the  same  altitude. 

Theorem  XIX. 

416.  The  area  of  a  rectangle  is  expressed  hy  the 
product  of  its  base  and  altitude. 

Hypothesis.    ABGD,  a  rectangle;  X,  the  unit  of  area. 

D, ^   P  Q 


BiT 


N 


Conclusion.  When  AB  and  CD  are  exprtssed  in  numbers 
of  which  the  side  of  X  is  the  unit,  then  the  area  ABCD  is 
expressed  in  units  by  the  product 

AB  X  AD. 
Proof.     Let  us  put 

a,  the  number  of  units  in  AB; 
bf  the  number  of  units  in  AD^ 
that  is  (§  352,  II.),  a  =  ratio  AB  :  side  X, 

b  =  ratio  AD  :  side  X. 
The  numbers  a  and  b  may  be  either  entire,  fractional,  or  in- 
commensurable. 

Construct  a  rectangle  MNPQ,  having  MN  =  side  X  and 
MP  =  b.    Consider  MP  as  a  base  of  this  rectangle.  Then— 
1.  Because  MN  =  side  X, 

Area  MNPQ  :  X  ..  MP  -.  side  X; 
that  is.  Area  MNPQ  :  X  =  b. 

linrw      TLtD   A    r» 

iUOC    JSLJ.       -_i    uXL/f 

ABCD  :  area  MJSTPQ  ::  AB  :  MN ::  AB  :  side  X; 
,  Area  ABCD  :  area  MJSTPQ  =  a. 


9.    R 


196         BOOK  V.     PROPORTION  OF  MAQNITUDES. 

Compounding  the  ratios  (2)  and  (1), 

Area  ABCD  \  X—  ah.  (§  373) 

That  is,  the  area  ABCD  is  expressed  by  ab  when  X  is  taken 
as  the  unit  (§  352).     Q.E.D. 

Corollary  1.  Because  a  triangle  has  one  half  the  area  of  a 
rectangle,  with  the  same  base  and  altitude,  we  conclude: 

417.  Th6  area  of  a  triangle  is  represented  by  one  half  the 
product  of  its  base  and  altitude. 

Cor.  2.  ItAB  =  AD,  then  a  =  b  and  ab  =  a\     Hence: 

418.  The  area  of  the  square  on  a  line  is  expressed  alge- 
braically by  the  square  of  the  number  of  units  in  the  line. 

We  thus  prove,  for  all  cases,  the  theorems  proved  for  whole  num- 
bers only  in  Book  IV.,  §§  284,  285, 

419.  Cor.  d.  The  areas  of  rectangles  or  triangles  having 
equal  bases  are  th  each  other  as  their  altitudes. 

Theorem  XX. 

420.  If  four  straigM  lines  are  proportional^  the 
rectangle  contained  hy  the  extremes  is-  equal  to  the 
rectangle  contained  hy  the  means. 

Hypothesis.    Four  straight  lines  o,  b,  c,  d,  such  that 

a  '.  b  '.:  c  :  d. 

Conclusion.    Rect.  a.d  =  rect.  b.c. 

Proof.  Form  the 

rectangles  ad  and  be,  , ^ , ^q 

and  place  them  so 
that  the  sides  d  and 
c  shall  be  in  one 
straight  line  and  the 
sides  a  and  b  in  an- 
other straight  line, 
crossing  the  first  at 
P.  Complete  rect- 
angle P^.  Then— 

1.  Because  the  rectangles  PQ  and  a.d  have  the  same  alti- 
tude a,  and  the  bases  c  and  d. 


a 

• 

e         I 

A        ■! 

d 

P 

% 

If—' 

ill—. 

A 

w 

Kect.  PQ  :  rect.  a.d  ::  c  :  d. 


(§413) 


ABEA3. 


107 


2.  Bacause,  taking  a  and  b  as  bases,  the  rectangles  PQ  and 
b.c  have  the  same  altitude  c,  and  the  bases  a  and  b, 

Eect.  FQ  :  rect.  b.c  ::a  :b;  ' 

or,  because  a  :  J  ::  c  :  <?, 

Rect.  PQ  :  rect.  J.c  ::  c  :  rf. 

3.  Comparing  with  the  proportion  (1);  three  terms  are 
found  equal.    Therefore  the  fourth  are  also  equal,  and 

Rect.  a.rf=  rect.  5.C.    Q.E.D. 

Scholium.  This  theorem  corresponds  to  the  theorem  of 
algebra  that,  if  four  numbers  are  proportional,  the  proc'^jt 
of  the  extremes  is  equal  to  the  product  of  the  means. 

Corollary.   If  ABGand  PQR  be  two  similar  polygons  in 
which  the  sides  ABy  BC, 
and     (JA    are    respectively 
homologous  to  PQ^  QR,  and 
RPy  we  shall  have 

AB  :  BO  ::  PQ  :  QR, 
and  therefore 

Rect.^^.  QR=rG(ii.BapQ. 
Hence: 

421.  In  two  similar  po- 
lygons the  rectangle  formed 
by  any  side  of  the  one  and  any  side  of  the  other  is  equal  in 
area  to  that  formed  by  the  homologous  sides. 

Theorem  XXI. 

432.  Converself/,  if  two  rectangles  are  eqtml  in 
area,  the  sides  of  the  one 

will  form  the  extremes,    ■*"" 

and  the  sides  of  the  other 
the  means,  of  a  propor- 
tion. ▲ 

Hypothesis. 
Rect.  ABCD^xQGi.  PQRB, 

Conclusion. 

AB  •  BR  ::  PB  :  BG, 

Proof.     Place   the    .     t- 
angles  so  that  the  sides  AB  and  BR  shall  be  in  one  straight 


B 


I 

I 
I 
I 
I 

R 


a 


198 


BOOK  V.    PROPORTION  OF  MAQNITUDES, 


line  and  the  sides  PB  and  BC  in  another  straight  line,  and 
complete  the  rectangle  BR8C.    Then — 

1.  Because  the  rectangles  ^(7  and  PR  are  equal, 

Rect.  AC  :  rect.  ^*S^  ::  rect.  PR  :  rect.  B8. 

2.  Because  the  rectangles  AO  and  BS  have  the  same 
altitudes  BC,  and  stand  on  the  bases  AB  and  BR, 

Rect.  AC  :  rect.  BS  ::  AB  :  BR. 

3.  In  the  same  way, 

Rect.  PR  :  rect.  BS  ::  PB  :  BG, 

4.  Comparing  with  (1)  and  (2), 

AB  :  BR  ::  PB  :  BC.     Q.E.D. 


Theorem  XXII. 

423.  The  areas  of  similar  triangles  are  to  each 
other  as  the  squares  of  their  homologous  sides. 


Hypothesis.   ABC  and  PQR,  two  triangles  in  which 

AB  '.BC:  CA  ::  PQ  :  QR  :  RP. 
Conclusion.   Area  ABC  :  PQR  ::  AB'  :  PQ\ 
Proof.   Construct  a  third  triangle  if,  of  which  the  base 

shall  be  equal  to  PQ,  and  the  altitude  to  CD,  the  altitude  of 

the  triangle  ABC.     Then — 

1.  Because  the  triangles  M  and  ABC  have  the  same  alti- 
tude CD,  they  are  to  each  other  as  their  bases  (§  414).  There- 
fore Area  ABC  :  area  M  ::  AB  :  PQ. 

2.  Because  the  triangles  M  and  PQR  have  equal  bases, 
they  are  to  each  other  as  their  altitudes.     Therefore 

Area  M  :  area  PQR  ..CD:  RS.  {§  419) 

3.  In  the  triangles  CAD  and  RP8  we  have 

Angle  A  =  angle  P  (by  hypothesis). 
Angle  D  =  angle  S  (right  angles). 


AREAS. 


199 


le  same 


Therefore  angle  ACD  ^  angle  PRS,  and  the  two  trianriea 
are  similar,  so  that 

CD  :  ES  ::  CA  :  RP  ::  AB  :  PQ, 
Therefore,  from  (2), 

Area  M  :  area  PQR  ::  AB  :  PQ. 
4.  Compounding  the  ratios  (1)  and  (3),  we  haye 
Area  ABC  :  area  PQR  ::  AB*  :  PQ\     Q.E.D. 

Corollary.  The  preceding  theorem  may  be  expressed  in 
the  following  form: 

434.  The  ratio  of  the  areas  of  two  similar  triangles  is 
the  square  of  the  common  ratio  of  each  side  of  the  one  to  the 
homologous  side  of  the  other. 

Theorem  XXIII. 

425.  The  areas  of  similar  polygons  are  to  each 
other  as  the  squares  v/pon  their  homologous  sides. 

Hypothesis.   ABCD  and  A'B'G'D',  two  similar  polygons, 
in  which  ^^  and  ^'^' are 
homologous  sides. 

Conclusion. 
A.vQdLABCD'.si,YeiiA'B'C'D'    D         ^ 
:iAB^'.A'B'\ 

Proof     Let  the  poly- 
gons be  divided  into  simi-    ^ 
lar  triangles  by  lines  drawn  ^ 
from^  (Th.  IX.).     Then— 

1.  Because  the  triangles^i5(7and^'i?'(7'are  similar,  and 
the  sides  AB  and  A'B'  are  homologous. 

Area  ABC  :  area  A'B'C  ::  AB^  :  A^B'\ 
3.  In  the  same  way  we  have 

Area  ABC  :  area  A'D'C'  ::  AD*  :  A'D'*- 
or,  because  AD  and  AD',  as  well  as  ^i5  and  A'B\  are  homol- 
ogous. 

Area  ADC  :  area  A'D'C  ::  AB^  :A 'B'\ 
3.  Continuing  these  proportions  through  the  whole  poly- 
gon, and  then  adding  all  the  antecedent*  and  conseauents 
we  have  (§371) 

Area  ABCD  :  area  A'B'G'D'  ::  AB'  :  A'B'\     Q.E.D. 


■1 

1 

■  ; 

f 

^^^H  ** 

■- 

^^B . 

1^ 

f 

' 

W[ 

J 

ii 

goo         BOOK  V.    PROPORTION  OF  MAQNITUDSS, 

Corollary  1.  The  result  of  this  theorem  may  bo  expressed 
in  the  following  form: 

426.  The  ratio  of  the  areas  of  similar  polygons  is  the 
square  of  the  ratio  of  each  side  of  the  one  to  the  homologous 
side  of  the  other. 

427.  Cor.  2.  If  three  lines  form  a  proportion,  the  area  of 
a  polygon  upon  the  first  is  to  the  area  of  the  similar  polygon 
upon  the  second  as  the  first  is  to  the  third. 

Also,  the  areas  of  the  similar  polygons  upon  the  second  and 
third  lines  are  in  the  same  ratio. 


Theorem  XXIV. 

428.  ff  two  chords  in  a  circle  cut  each  other ^  the 
rectangle  of  the  segments  of  the  one  is  equal  in  area 
to  the  rectangle  of  the  segments  of  the  other. 

Hypothesis.  AB  and  CD,  two  chords  intersecting  at  P. 
Conclusion. 

Kect.  AP.PB  =  rect.  CP.PD. 
Proof.    Join  AD  and  BC.    Then— 
1.  Because    the    angles    ADO  and  ^/ 
ABC  stand  on  the  same  arc  AC, 

Angle  ADC  =  angle  ABC.  (§  237) 
Therefore,  in  the  triangles  APD  and 
BPC,  we  have 

Angle  APD  =  opp.  angle  BPC. 
Angle  ADP  =  angle  PBC  (on  same  nrc  /I  C), 
Angle  PAD  =  angle  PCB  (on  samo  aio  BD). 
Therefore  these  triangles  are  similar  (§  390),  and  the  sides 
opposite  the  equal  angles  are  proportional,  so  that 

AP  :  PD  ::  CP  :  PB. 
%    r>eeat,use  of  this  proportion, 

B'   t.  AP.PB  =  rect.  CP.PD  (§  420).    Q.E.D. 

Scholium.  This  theorem  and  the  corollary  of  the  following 

atxj    iwciinvjai    micii    ttu    v>OiiDi«.ci.     liiic    v;iiuiui3    us    Cutting    c»uii 

other  externally  when  they  do  not  meet  within  the  circle. 


AJUSAS. 


1 
201 


Theorem  XXV. 

429.  I/from  a  point  witJiout  a  circle  a  secant 
and  tangent  be  drawn,  the  rectangle  of  the  whole 
secant  and  tJie  part  outside  the  circle  is  equal  to  the 
square  of  the  tangent. 

Hypothesis,  P,  a  point  with- 
out a  circle;  PT,  a  tangent  touch- 
ing the  circle  at  T;  PB,  a  secant 
cutting  the  circle  at  A  and  B. 

Conclusion.  PA.PB  =  PT\ 

Proof.  Join  TA  and  TB. 
Then— 

1.  Because  the  angle  ^^T  stands  upon  the  arc  TA, 

Angle  ABT=i  angle  arc  A  T.  (§  235) 

2.  Because  PT'i^  a  tangent,  and  ^T a  chord. 

Angle  PTA  =  \  angle  arc  A  T.  (§  234) 

3.  Therefore  angle  ABT^  angle  PTA,  and  the  triangles 
P^Tand  PTB  have  the  angles  at  P  identical  and  two  other 
angles  equal.  Therefore  the  third  angles  PTB  and  PA  T are 
also  equal,  and  the  triangles  are  similar  (§  390). 

4.  Comparing  homologous  sides,  we  have 

PA  :  PT::  PT :  PB, 

5.  Therefore 

Rect.  PA.PB  =  PT'  (§  420).    Q.E.D. 

430.  Corollary.  Because 
the  rectangles  formed  by  all 
secants  from  P  are  equal  to  the 
square  of  the  same  tangent 
PT,  they  are  all  equal  to  each 
other. 

Theorem  XXVI. 

431,  When  the  bisector  of  an  angle  of  a  triangle 
meets  the  base,  the  rectangle  of  the  two  sides  is  equal 
10  the  rectangle  of  the  segments  of  the  base  plus  the 
square  of  the  bisector. 


302         BOOK  V.     mOPOHTIOI/  OF  MAGmTl.DE8.      ' 

Hypothesis.    ABC,  any  triangle;  CD,  the  bisector  of  the 
onglo  at  (7,  cutting  the  base  at  i>. 

Conclusion,  \ 

Reot.  CA,  CB  =  rect.  AD,DB  +  CD\  -''""x^X 

Proof,      Circumscribe     a    circle       /^  y^'^x  \N 
A  OBE  «.roui?d  the  given  triangie,  and      ;[y^       \        \ 
continue   the  bisector  till    it   meets  ^\  fe      yl* 

the  circle  in  E,    Join  BE.    Then—        \ 

1.  In  the  triangles  CaD  and  GEB      \ 
we  have  '"-...  _ 

Angle  AGD"  angle  BCE  (hyp. ).  "^ 

Angle  CAD  =  angle  BEC  (on  same  ore  BC), 
Therefore  these  triangles  are  equiangular  and  similar. 

2,  Comparing  the  sides  opposite  equal  angles, 

,     CA  :  CD  i:  CE  :  CB. 
Whence 

Kect.  CA.CB  =  rect.  CE.CD, 

=  Teot{CD-\-DE)CD, 

=  CD'  +  rect  CD.DE,  (§  287) 

=  CD*  +  rect.  AD.DB  (§  428).    Q.E.D. 


Theorem  XXVII. 

432.  M  a  riglit-angled  triangle  the  area  of  any 
polygon  upon  the  hypothenme  is  equal  to  the  sum  of 
the  areas  of  the  similar  and  similarly  described 
polygons  upon  the  two  other  sides. 

Hypothesis.  ABC,  a  triangle,  right-angled  at  A.  We  also 
put 

a,  a',  a", 

the  areas  of  any  three  similar 
and  similai-ly  described  poly- 
gons, upon  the  sides  BC,  CA, 
and  ABy  respectively. 

a  =  a  -j-  a 


i^OflCiUSlOil. 


Proof.    From  A  drop  AD  ±  BC.    Then- 


PBOBLEMB  IN  PROPORTION. 


203 


1.  Because  ABC  is  right-angled  at  A,  and  AD  is  a  per- 
pendicular upon  the  hypothenuse, 

DC  :  CA  ::  CA  :  BC.  ' 

DB  :  BA  ::  BA '.  BG, 

2.  Because  a\  a",  and  a  are  the  areas  of  similar  poly- 
gons described  upon  CA,  BA,  and  46", 

a'   :a::  DO  :  BC. 

a":a::BD:BC.  (§427) 

3.  Therefore,  taking  the  sum  of  the  ratios  (§  366), 

a'  +  a"  :  a  ::   BD-{  DC  :  BC. 

4.  Because  BD-\-DC=BC,  ibhe  second  ratio  is  unity; 
therefore  the  first  also  is  unity,  and 

«'  +  «"  =  «.     Q.E.D. 
Scholium.  This  result  includes  the  Pythagorean  proposi- 
tion  (§308),  as  a  special  case  in  which  the  polygons  are 
squares. 


»  »  > 


CHAPTER   IV. 

PROBLEMS    IN    PROPORTION. 


A— 


Problem  I. 

433.  To  divide  a  straight  line  similarly  to  a 
given  divided  straight  line. 

Given.  Aline,  AB-,  another  line,C7/>,  divided  at  the  points 

M  und  N. 

Required.     To  divide  AB  t 
similarly  to  CD. 

Analysis.  By  §388  two 
sides  of  a  triangle  are  similarly 
divided  by  any  lines  parallel  to 
the  base.  Therefore,  if  we  put 
together  the  lines  AB  and  CD 
m  such  a  way  as  to  form  two 

siflofl    /-if    o     +«;«».^i^     «n     i: , 

parallel  to  the  third  side  will 
divide  these  two  sides  similarly. 


r 


M 


illi 


204         BOOK  V.    PROPORTION  OF  MAQNITUDEa. 

Construction.   1.  Form  a  triangle  ABD,  such  that 
AB  =  given  line  AB, 
AD  =  given  line  CD,  \ 

BD  =  any  convenient  length. 
2.  Through  the  points  M  and  N  draw  MM'  and  NN' 
parallel  to  BDy  meeting  AB  in  M'  and  N'.     Then 

AM' '.M'N' :N'B',:AM:MN',ND.       (§388) 

Therefore  the  line  ^5  is  divided  at  the  points  M'  m^N' 
similarly  to  CD,    Q.B.F. 

Problem  II. 

434.  To  divide  a  straight  line  internally  into 
segments  which  shall  be  to  each  other  as  two  given 
straight  lines.  p 

Given.  Two  straight  lines,  y 
p  and  q ;  a  third  straight 
line,  AB, 

Required.  To  divide  AB 

into    segments   having    the       A'^ ^ ^B 

same  ratio  2ii&  p  to  q. 

Construction.  1.  From  one  end  of  the  line  AB  draw  an 
indefinite  straight  line  AD. 

2.  From  this  line  cut  off  ^  C  =  ^  and  CD  =  q. 

3.  3om  DB. 

4.  From  C  draw  a  line  parallel  to  DB,  and  let  E  be  the 
point  in  which  it  cuts  AB. 

The  line  AB  will  be  cut  internally  at  E  into  the  segments 
AE  and  EB^  having  to  each  other  the  same  ratio  as  the  lines 
p  and  q. 

Proof.    As  in  Problem  I. 

Problem  III. 

435.  To  divide  a  straight  line  externally  so  that 
the  segments  shall  he  to  each  other  as  two  given 
straight  lines. 

/3V^./.*.      A   a4-T.oir»lif  lirjo     A  Ti'  fwo  nfliP.r  linfis.  «,  «, 

Required.  To  cut  AB  externally  so  that  the  segments 
shall  have  to  each  other  the  ratio  p  :  q. 


PROBLEMS  IN  PROPORTION. 


206 


Construction.    1.  From  either  end  of  the  line  AB  as  A 
draw  an  indefinite  straight  line  ^C.  '         ' 

2.  On  this  line  cut  oft  AC  =  ; 

the  greater  line  p,  and  from  C,  ^ — ~~ 

toward  A,  cut  off    CD  =  the  * ^<^ 

lesser  line  q.  ^<^^'^^^^^^^ 

3.  Join  DB.  ^^^^^  \ 

4.  From  C  draw  a  line  par-     .^^^^         \       _^\ 
allel  to  DB,  and  let  B  be  the  B  "  E 
point  in  which  it  cuts  AB  produced. 

The  line  AB  will  be  divided  externally  at  B,  so  that 

AH  :  BU  ::p  :  q. 

Proof.    As  in  Problem  I. 

436.  Corollary  1.  If  ^  =  g,  the  point  D  would  fall  upon 
A,  and  the  line  DB  would  coincide  with  ^^.  The  line  drawn 
through  C  parallel  to  DB  would  then  be  parallel  to  AB,  so 
there  would  be  no  point  of  intersection  B.  Therefore  there 
would  .be  no  external  point  for  which  the  ratio  of  the  see- 
ments  would  be  unity.  ^ 

43 Y.    Cor.  2.  If  we  combine  Problems  11.  and  III  on 

the  same  straight  line,  using  the  same  ratio,  we  shall  divide 
the  line  harmonically,  and  the  ends  of  the  line,  together 
with  the  points  of  division,  will  form  four  harmonic  points. 

Problem  IV. 

438.  To  find  a  fourth  proportional  to  three  given 
straight  lines.  

Given.    Three  straight  lines, 
a,  h,  c. 

Required.    To  find  a  fourth 
line,  X,  such  that 

a  :  b  ::  c  :  X. 
Analysis.    To  solve  the  prob- 
lem it  is  only  necessary  to  form 

two  similar  triangles,  one  of  which     /                         ^\ 
shall  have  a  and  n  for  f.wn  r.f  i+«  ^  " —  ^^ 

sides,  while  the  other  shall  have  b  as  the  side  homologous 
to  a,    The  side  bomologovis  to  c  win  then  be  x. 


3 


\  !f 

n 

, 

4  ^H 

<t  m 

i  ma 

' 

m 

a- 
b- 
e. 


306  BOOK  V.    PROPORTION  OF  MAGNITUDES. 

Construction.     1.   Draw  two  straight  lines  from  the  same 

point  A, 

3.  From  one  of  them  cut  off 
AB  =^a  and  AD  =  b,  and  from 
the  other  cut  oft  AG  =  c. 

3.  Join  BO, 

4.  Through!)  draw  2)^11  i5a 

AE  will  be  the  required  line  x. 

Proof,  The  similar  triangles 
^ J?(7  and  ^/>j^  give 

AB  '.AD  ::AG:AE, 
which  is  the  required  proportion. 

Problem  V. 
439.    To  find  a  rnean  proportional  between  two 
given  straight  lines. 

Given.  Two  straight  lines,  AD  and  DB, 

Required.  To  find  a  third  line  which  shall  be  a  mean 
proportional  between  them.  a D 

Analysis,  The  perpendicular  from 
any  point  of  a  circle  upon  the  diam- 
eter is  a  mean  proportional  between 
the  segments  into  which  it  divides 
the  diameter  (§  402).     Hence 

Construction.  1.  On  an  indefinite 
line  take  the  segments  AD  and  DB^ 
equal  to  the  given  lines. 

2.  On  ^  5  as  a  diameter  describe  a  semicircle. 

3*.  At  D  erect  the  perpendicular  DC,  meeting  this  circle  in 
C.    DC  will  then  be  the  required  mean  proportional  between 

ADmdDB.  ^      ,    .    ^.        ceonQ 

Proof,  This  may  be  supplied  by  the  student  from  §§^0S 

and  401. 

440.  Def  A  straight  line  is  said  to  be  divided  in 
extreme  and  mean  ratio  when  it  is  divided  into  two 

+«    aiinh  fhnt  thft  crreater  sesrment  is  a  mean 

proportional  between  the  lesser  one  and  the  whole 
line. 


-B 


-.0 

- —     r^x-^ 


a 


\ 


s 

\ 


D     b 


B 


PROBLEMS  IN  PROPORTION, 

Problem  VI 


207 


441.   To  divide  a  straight  line  in  extreme  and 
mean  ratio. 

Construction.    Let  AB  be  the  given  straight  line.  Then— 

1.  At  one  end  B  of  the  given 
straight  line  erect  a  perpendicular 
BOy  and  take  BO  =  ^AB, 

2.  Around  Cas  a  centre,  with  a 
radius  CB  describe  a  circle. 

3.  Join  AC,  produce  the  line 
A  Cy  and  let  ^  and  i>  be  the  points  a- 
in  which  it  intersects  the  circle. 

4.  From  ^  as  a  centre  draw  an  arc  cutting  off  from  AB 
the  length  ^^  =  ^^.  ^       "um  ^zj 

The  Une  AB  will  then  be  cut  at  i^in  such  manner  that 
FB  :  AF ::  AF '.  AB; 
that  IS,  it  wiU  be  cut  at  i^in  extreme  and  mean  ratio 

Proof,   1.  Bemuse  AB  is  perpendicular  to  the  radius  CB 
at  Its  extremity,  B,  it  is  a  tangent  to  the  circle.    Therefore 

2.Bydivision,^^^^^--^^^^^^-  (§^^9^  4) 

AE:AB-AEi:AB:AD-^AB,  (§365) 

An      r^    ^"^       ^"^  ^^^  diameter  ED,  so  that  AB  -  AB  = 
An^ED=  AE  =  AF.    Making  these  substitutions  in  (2) 

.  AF:FB::AB:AF;  ^  ^' 

or,  by  inversion,  ' 

FB  :AF::AF:AB. 

4.  Therefore  the  segment  AF  is  a  mean  proportional 

between  the  segment  FB  and  the  whole  line  AB      ^     "^""^ 

443.  Scholium.    This  division  of  the  straight  line  has 

take  i^(7  =  FB.  the  linfi  /4  P  win  v>«      iT  o   A  d      ^ 
cut  m  extreme  and  mean  ratio 


FB  :  AF ::  ^j^ ;  J^ 


6^.    For,  by  hypothesis. 


Il  III 

1    <\ 


208         BOOK  V.     PROPORTION  OF  MAQNITUDEB, 

whence,  by  inversion  and  diyision, 

AF-  FB  :  FB  ::  AB  -AF:  AF, 
Because  OF  =  FB,  this  proportion  is  the  same  as 

AG  :  GF::  GF :  AF. 
In  the  same  way,  by  taking  on  GF  a  line  Gh  equal  to  -4  G^ 
we  shall  divide  GF  in  extreme  and  mean  ratio  at  h,  and 
so  on  indefinitely. 

443.  Corollary.  The  two  segments  formed  ly  the  **  golden 
section"  are  incommensurable  with  each  other. 

For  suppose  ^i^were  composed  of  m  parts,  and  FB  of  n 
equal  parts.     Then  when  we  cut  off  FG,  we  should  have 

AG  =  m  —  n  parts. 
Cutting  this  off  from  GF,  we  should  have  in  hF  a  still  smaller 
number  of  parts.     But  no  part  would  ever  be  divided  by  cut- 
ting, because  when  we  subtract  one  whole  number  from  an- 
other, the  remainder  is  a  whole  number. 

Therefore,  because  every  time  we  cut  off  we  reduce  the 
number  of  parts  by  one  or  more,  our  last  section  would  take 
away  all  the  line  that  was  left,  and  so  would  not  cut  it  in 
extreme  and  mean  ratio.  |   '■}'■(>  |   i   ■■■   | 

lUmtraUon.    If  AF  had  eight  ^         O  A      F  B 

parts,  and  FB  five  parts,  then  by  successively  cutting  off  we  should 

have  left 

AQ  =  Z  parts  =  Qh. 

hF  =  2  parts. 

Cutting  off  these  two  parts  from  Oh,  we  should  have  one  part  left,  and 

after  two  more  cuttings  nothing  would  be  left. 

The  Phyllotaxis. 
444.    Let  us  bend  the  line  AB  around  into  a  circle, 
the  two  ends,   A  and  B,   coming  together.      Let  us  then 
take  the  distance  BF  in  our  dividers, 
and,  starting  from  the  point  AB,  keep 
measuring  off  equal  steps  round   and 
round  the  circle.     The  end  of  each  step 
is  numbered  from  0  through  1,  2,  3,  etc. 
Then— 

1.  At  every  step  we  shall  find  the 
circle  divided  into  arcs  of  two  or  three 
different  lengths. 


PROBLEMS  IN  PROPORTION. 


209 


2.  At  every  step  the  dividers  will  fall  upon  one  of  the 
longer  arcs,  and  will  divide  it  in  extreme  and  mean  ratio. 

3.  The  division-marks  will  be  scattered  around  the  circle 
more  evenly  than  by  any  other  system  of  division. 

It  has  been  considered  by  botanists  that  the  leaves  of 
plants  are  arranged  around  the  stem  on  such  a  plan  as  this. 
This  arrangement  is  then  called  the  phyllotaxis. 

Problem  VII. 

445.  Upon  a  gimn  straight  line  to  construct  a 
polygon  similar  to  a  gimn  polygon. 

Given.  A  polygon, 
ABODE)  a  straight 
line,  PQ. 

Required.  To  con- 
struct upon  PQ  a  poly- 
gon similar  to  ABODE. 

Oonstruction.  Let 
AB  be  the  side  of  the  given  polygon  to  which  PQ  is  to  be 
homologous.  Divide  the  given  polygon  into  triangles  by  diaff- 
onals  from  the  point  A. 

UponP^  describe  the  triangle  PQR  equiangular  to  ^  J? a 

Upon  PR  describe  the  triangle  PRS  equiangular  to  A  OD, 
and  continue  the  process  through  all  the  triangles  into  which 
^5Ci)^  is  divided. 

The  polygon  PQRSTmW  be  the  one  required. 

Proof.    By  §395. 

Problem  VIII. 

446.  To  describe  a  polygon  wMch  shall  he  equal 
to  one  and  similar  to  the  other  of  two  given  polygons. 

Given.    Two  polygons,  P  and  Q. 

Required.  To  construct  a  third  polygon,  equal  in  area 
to  P  and  similar  to  Q. 

Analysis.  Because  the  required  figure  is  similar  to  Q,  the 
square  of  any  one  of  its  sides  will  have  the  same  ratio  to  thfi 
square  of  the  homologous  side  of  Q  that  its  area  has  to  the 
area  of  Q. 


ii 


210         BOOK  V.    PROPORTION  OF  MAGNITUDES. 

Because  the  area  is  that  of  P,  this  ratio  of  the  squares 
of  homologous  sides  is 
the  ratio  of  the  area  of 
P  to  the  area  of  Q. 

Therefore  if  we  con-       /  p 

struct  two  squares,  the 
one  equal  in  area  to  P 
and  the  other  to  Q,  the 
sides  of  these  squares  will  have  the  same  ratio  which  each 
side  of  the  required  figure  has  to  the  homologous  side  of  Q. 

Construction.     Construct  the  side  rf  of  a  square  equal  in 
area  to  P,  and  the  side  g  of  another  square  d 

equal  in  area  to  Q  (§  321). 

Take  any  side  MN  of  Q  and  find  a  fourth 
proportional  h  to  g,  d,  and  MN  (§  438). 


h 


Upon  h  describe  a  polygon  X  similar  to  ^,  and  having  l 
as  the  side  homologous  to  MN. 

This  polygon  X  will  be  equal  in  area  to  P,  as  well  as 

similar  to  Q. 

Proof.  Because  h  and  MN  are  homologous  sides  of  the 
similar  polygons  X  and  Q, 

Area  X  :  area  Q  ::  h^  :  MN*; 
or,  because  by  construction  h  :  MN  ::  d  :  g, 
Area  X :  area  Q  ::  d*  :  g\ 
But,  by  construction,  d"  =  area  P  and  g*  =  area  Q, 
Therefore  Area  X  :  area  Q  ::  area  P  :  area  Q. 

Whence  Area  X=  area  P.     Q.E.F. 

Theorems  for  Exercise. 

Theobem  1.  If  the  ends  of  two  intersecting  chords  be 
joined  by  straight  lines,  the  two  triangles  thus  formed  will  J)e 
similar  to  each  other. 

Theorem  2.  If  two  chords  of  circles  subtend  arcs  which 
together  make  up  a  semicircle,  the  sum  of  their  squares  is 
equal  to  the  square  of  the  diameter. 

Theorem  3.  If  from  any  point  within  a  parallelogram 
lines  be  drawn  to  the  four  vertices,  the  sum  of  the  areas  of 


THEOREMS  FOR  EXERCISE. 


211 


each  pair  of  opposite  triangles  is  half  the  area  of  the  paral- 
lelogram. ^    " 

Theorem  4.     If  two  equal    tri- 
angles on  the  same  base  be  cut  by  .  a/..\b...Bk 
a  line  parallel  to  the  base,  equal  areas 
will  be  cut  off  from  them. 
Area  ABC  =  DEF. 

Theorem  6.  Lines  drawn  through  the  point  of  contact  of 
two  circles,  and  terminated  by 
the  circles,  form    four  chords 
which  are  proportional,  and  the  \ 
lines  through  the  ends  of  which  ^| 
are  parallel. 

CoTidusions, 

I.    AP  :  PC  ::  BP  :  PD. 

II.    ACWDB. 

Theorem  6.  If  a  circle  be  described  touching  two  paral- 
lels, and  from  the  points  of  tan-  ^ 
gency  secants  be  drawn  intersect- 
ing the  circle  in  the  same  point, 
and  terminated  by  the  opposite 
parallel,  the  diameter  of  the  circle 
will  be  a  mean  proportional  be- 
tween the  segments  of  the  paral- 
lels. 

HypotTism.  The  lines  PB  and  QA  intersect  at  B,  a  point  of  the  circle 
Condusion.  BQ  :  PQ  ::  PQ  :  AP. 

Corollary.  The  same  thing  being  supposed,  prove 
Kect.  RA.RB  =  rect.  RP.RQ, 

Theorem  7.  If  through  any  ver-Gr-  ^ 

tex  of  a  parallelogram  a  line  be  \ 
drawn  meeting  the  two  opposite 
sides  produced  without  the  paral- 
lelogram, the  rectangle  of  the  pro- 
duced portion  of  such  sides  is  equal 
to  the  rectangle  of  the  sides  of  the  \ 

'ixl  tliiciugi  aiii.  \  ^' 

Eypothem.   OD.  BF=  AB.  BC.  ' 


I 


212         BOOK  V.     PROPORTION  OF  MAGNITUDES. 

Theorem  8.  If  two  triangles  have  one  angle  of  the  one 
equal  to  one  angle  of  the  other, 
and  the  perpendiculars  from  the 
other  two  angles  upon  the  oppo- 
site sides  proportional,  they  are 

similar. 

Jlypothem.  Angle  0  =  angle  C 

AP.BQr.A'P  :BQ'. 
Condumn.  The  triangles  are  similar. 

Theorem  9.  If  the  four  sides 
of  an  inscribed  quadrilateral  taken  consecutively  form  a  pro- 
portion, the  diagonal  having  the  means  on  one  side  and  the 
extremes  on  the  other  side  divides  it  into  two  triangles  of 
equal  area.     (Book  IV.,  Ex.  Th.  9.) 

Theorem  10.  The  rectangle  of  two  sides 
of  a  triangle  is  equal  to  the  rectangle  of  its 
altitude  above  the  third  side  and  the  diam- 
eter of  the  circumscribed  circle. 

Theorem  11.  The  area  of  a  triangle  is 
equal  to  the  product  of  the  three  sides  divided  by  twioo  the 
diameter  of  the  circumscribed  circle. 

Theorem  12.  If  two  parallel  tangents  to  a  circle  are  inter- 
cepted by  a  third  tangent,  the  rectangle  of  the  segments  of 
the  latter  is  equal  to  the  square  of  the  radius  of  the  circle. 

Theorem  13.  If  a  chord  be  drawn  parallel  to  the  tangent 
at  the  vertex  of  an  inscribed  triangle,  the  portion  of  the  tri- 
angle cut  off  by  the  chord  is  similar  to  the  original  triangle. 

Theorem  14.  If  from  the  middle  point  A  of  the  arc  sub- 
tended by  a  fixed  chord  a  second  chord  be  drawn  inter- 
secting the  fixed  one,  the  rectangle  contained  by  the  whole 
of  that  second  chord  and  the  part  of  it  intercepted  between 
the  fixed  chord  and  the  point  it  is  a  constant  whatever  be  the 
direction  of  the  second  chord. 

Show  to  what  square  or  area  the  constant  area  is  equal. 

Theorem  15.  If  in  two  triangles  any  angle  of  the  one  is 
equal  to  some  angle  of  the  other,  their  areas  are  to  each  other 
as  the  rectangles  of  the  sides  which  contain  the  equal  angles. 


NUMERICAL  EXEBCISEa. 


213 


Theorem  16.  If  two  chords  of  a  circle  intersect  each 
other  at  right  angles,  the  sum  of  the  squares  of  the  four 
segments  is  equal  to  the  square  of  the  diameter. 

Theorem  17.  If  on  each  of  the  sides 
of  an  angle  having  its  vertex  at  0  two 
points  A  and  B  on  the  one  side  and  P 
and  Q  on  the  other  be  taken  such  that 

OP  :  OA  ::  OB  :  OQ, 
the  four  points  A,  B,  P,  and  Q  will  lie 
on  a  circle  (§§  244,  392). 

Theorem  18.  If  at  any  point  outside 
of  two  circles  a  point  be  chosen  from  which  the  tangents  to 
the  two  circles  shall  be  equal  in  length,  and  from  this  point 
secants  to  each  circle  be  drawn,  the  four  points  of  intersection 
will  lie  on  a  third  circle. 

Apply  §  429,  4,  to  the  case  of 
each  circle. 

Theorem  19.  Conversely, 
if  two  circles  be  intersected 
by  a  third,  and  secants  be 
drawn  through  each  pair  of 
points  of  intersection,  the 
tangents  to  the  circles  from 
the  point  of  intersection  of 
the  secants  will  be  equal  in  length. 

Theorem  20.  If  the  common  secant  of  two  intersecting 
circles  be  drawn,  the  tangents  to  the  two  circles  from  each 
point  of  this  secant  will  be  equal  in  length. 

NuMEEicAL  Exercises. 

1.  If  one  of  two  similar  triangles  has  its  sides  50  per  cent 
longer  than  the  homologous  sides  of  the  other,  what  is  the 
ratio  of  their  areas  ? 

2.  The  owner  of  a  rectangular  farm  containing  10,000 
square  yards  finds  that  it  measures  5  inches  X  20  inches  on  a 
map.     What  are  its  length  and  breadth  ? 


II 

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I 

BOOK  VI. 

REGULAR   POLYGONS  AND   THE 

CIRCLE. 


CHAPTER   I, 


PROPERTIES  OF   INSCRIBED  AND   CIRCUMSCRIBED 
REGUUR   POLYGONS. 


Theorem  I, 


447.  If  a  circle  he  dimded  into  any  numher  of 
equal  arcs^  and  a  chord  he  drawn  in  each  arc,  these 
chords  will  form  a  regular  polygon. 

Hypothesis.  A,  B,  0,  D,  E,  equidistant  points  around  a 
circle,  separating  it  into  equal  arcs;  AB,  ^ 

BGf  etc.,  the  chords  of  those  arcs. 

Conclusion.    The  polygon -4 -BC/)^  is 
regular  (§  152). 

Proof.  I.     Because  the  sides  are  by  ^\ 
hypothesis  all  chords  of  equal  arcs,  they 
are  all  equal  to  each  other. 

II.  Take  any  two  angles  of  the  poly-  D 

gon,  say  ^^Cand  CDE.    Join  ^Cand  GE.    Then— 

2.  Because  the  arcs  A  C  and  GE  are  equal,  being  sums  of 

equal  arcs, 

Chord  AG^  chord  GE.  (§  208) 

3.  Hence  in  the  triangles  ABG  and  GDE  we  have 

AB  =  GD, 

BG  =  DE, 

AG^GE. 

Therefore  these  triangles  are  identically  equal,  and 

Angle  ABG=  angle  GDE. 


UfaCBIBED  AND  CIIWUMaVRIBED  POLYGONS.      215 


4.  In  tho  sumo  way  it  may  bo  shown  that  any  other  two 
angles  of  tho  inscribed  polygon  wo  choose  to  take  are  equal. 

6.  Comparing  (1)  and  (4),  tho  polygon  is  shown  to  be 
regular  (§  152).     Q.E.D. 

Scholium.  Tho  equality  of  the  angles  of  the  polygon  may 
be  proved  with  yet  greater  elegance  by  showing  that  they  are 
all  inscribed  in  equal  segments. 

Theorem  II. 

448.  If  a  circumscribed  polygon  touch  a  circle 
at  equidistant  points  around  it,  it  is  regular. 

Hypothesis.  A  circumscribed  polygon 
whose  sides  touch  tho  circle  at  tho  equi- 
distant points  ABODE. 

Conclusion.  This  polygon  has  all  its 
sides  and  angles  equal. 

Proof.  Let  0  be  the  centre  of  the 
circle.    Join  OA,  OB,  etc.     Then — 

1.  Because  tho  intercepted  arcs  AB, 
BO,  etc.,  are  equal,  we  shall  have 

Angle  AOB=  B00=  ODD,  etc. 
Turn  the  figure  around  on  the  point  0  until  the  radius 
OA  coincides  with  the  trace  OB.    Then — 

2.  Because  of  the  equality  of  the  angles  A  OB,  BOO,  etc., 
OB  will  fall  upon  the  trace  00;  00=  trace  OD,  etc. 

3.  Because  the  radii  are  equal,  the  point  A  will  fall  on  B, 
B  on  (7,  etc. 

4.  Because  each  radius  is  perpendicular  to  the  tangent  at 
its  extremity,  each  side  will  fall  upon  the  trace  of  the  side 
next  following. 

5.  Therefore  each  point  of  intersection  will  fall  on  the 
trace  of  the  point  next  following. 

6.  Therefore  each  side  and  angle  is  equal  to  the  side  and 

angle  next  following,  and  the  polygon  has  all  its  sides  and 

angles  equal.    Q.E.I). 

Remark.  Another  demonstration  may  be  found  by  draw- 
in  or  liTiPa  frnm  f)  +.n  f.Vio  ancrloo  t\f  +lia  nrtWrrATi    onrl  rkT»rk-«rJr»rf  flio 

equality  of  all  the  triangles  thus  formed. 


Ill' 


lit'  I 


■i\ 


Il  J 


I  I 


216    BOOK  VI.     BEQULAR  POLYGOM  AND  THE  OIMCLE. 


Theoeem  III. 

449.  A  circle  may  be  inscribed  in  any  regular 
polygon^  or  circumscribed  about  it 

Proof.   I.  Let  ABODE  be  the  regular  polygon.    Bisect 
any  two  adjacent  angles  of  the  polygon, 
as  A  and  B,  and  let  0  be  the  point  of 
meeting  of  the  bisecting  lines.    Then — 

1.  In  the  triangle  AOB  the  angles 
OAB  and  OB  A  are  by  construction  the 
halves  of  the  equal  angles  EAB  and 

^^C(hyp.). 

Therefore  ihe  angles  are  equal  and 
the  triangle  is  isosceles. 

2.  Join  00.    In  the  triangles  AOB  and  BOO  we  haye 

BO  =  AB  {hjg.). 
OB  =  OB  (common  sidie). 
Angle  OAB  =  angle  OBC  (halves  of  equal  angles). 

Therefore  these  two  triangles  are  identically  equal,  and 
0(7r=  OB. 
Angle  00 B  =  angle  OB  A  =  I  angle  B  =  ^  angle  0. 

3.  In  the  same  way  it  may  be  shown  that  if  we  join  0  to 
the  other  angles  of  the  polygon,  the  triangles  thus  formed 
will  all  be  identically  equal.     Therefore 

OA  =  0B=00=  0D=:  OB. 

Hence  if  a  circle  be  drawn  around  0  as  a  centre  with  a 
radius  equal  to  either  of  these  lines,  it  will  pass  through 
all  the  points  A,  B,  O,  D,  B,  and  will  be  circumscribed 
around  the  polygon.     Q.E.D. 

II.  Let  Oa  be  the  perpendicular  from  0  upon  AB. 

4.  Because  the  triangles  OAB,  OBG,  etc.,  are  identically 
equal  (2),  the  perpendiculars  from  0  upon  AB,  BO,  etc.,  are 
equal  (§  175). 

Therefore  if  a  circle  be  drawn  around  0  as  a  centre,  with 
a  radius  Oa,  this  circle  will  also  pass  through  the  feet  of  the 
perpendiculars  dropped  from  0  upon  BO,  upon  CD,  etc. 

6.  Because  each  of  the  sides  AB,  BO,  etc.,  is  perpen* 


INSCRIBED  AND  CIRCXIMaCBIBBD  POLYGONS.     217 

dicular  to  a  radius  at  its  extremity,  it  is  a  tangent  to  the 
cipcie* 

Therefore  the  circle  is  inscribed  in  the  polygon.    Q.E.D. 

450.  Corollary  1.  The  inscribed  and  circumscribed 
circles  of  a  regular  polygon  are  concentric. 

4:51.  Cor,  2.  The  bisectors  of  the  angles  of  a  regular 
polygon  all  meet  in  a  point,  which  point  is  the  centre  of  both 
the  inscribed  and  circumscribed  circles,  and  is  equally  distant 
from  all  the  angles  and  all  the  sides  of  the  polygon, 

4:52.  Def.  The  common  centre  of  the  inscribed 
and  circumscribed  circles  is  caUed  the  centre  of  the 
regular  polygon. 

463.  Cor.  3.  The  perpendicular  bisectors  of  the  sides  of 
a  regular  polygon  all  pass  through  its  centre. 

454.  Cbr.  4.  If  lines  be  dra^vn  from  the  centre  of  a  regu- 
lar polygon  to  each  of  its  vertices,  the  polygon  will  be  divided 
into  as  many  identically  equal  isosceles  triangles  as  it  has 
sides. 

Theorem  IV. 

455.  All  regular  polygons  having  the  same  num- 
oer  of  sides  are  similar  to  each  other. 

Proof  1.  If  the  polygons  have  each  n  sides,  the  sum  of  all 
the  n  angles  of  each  is  equal  ton -2  straight  angles  (§  160). 

2.  Because  the  angles  of  each  polygon  are  equal,  each 
angle  of  each  polygon  is  the  nth  part  otn-2  straight  angles; 
that  IS,  the  angles  of  one  polygon  are  each  equal  to  the  angles 
of  the  other.  ^ 

3.  Because  the  sides  of  each  polygon  are  equal  to  each 
other  (hypothesis),  the  ratio  of  any  side  of  the  one  to  any  side 
of  the  other  is  the  same  whatever  side  be  chosen. 

Therefore  the  polygons  are  similar  (§  375).    Q.E.D. 

Theorem  V. 

456.  Regular  inscribed  and  circumscribed  poly- 
gons of  the  same  number  of  sides  may  be  so  placed 
that  their  sides  shall  be  parallel,  and  each  vertex  of 
the  one  on  the  same  radius  with  a  vertex  of  the  other. 


Ill  I 


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218    BOOK  VI.    REGULAR  POLYGONS  AND  THE  OIROLE. 

Hypothesis.    ABCDE,  an    inscribed    regular  polygon; 
A'B'C'D'E'f   a  circumscribed  poly- 
gon of  the  same  number  of  sides,  hav- 
ing the  side^'-B'  tangent  to  the  circle 
at  jT,  the  middle  point  of  the  arc  AB- 

Conclusions.    I.  A'B'  \\  AB ; 

B'C  II  BG,  etc. 

II.  The  vertex  ^'  is  on  the  radius 
OA  produced,  and  all  the  other  ver- 
tices, B'y  C",  etc.,  are  on  the  radii 
OBy  00 f  etc.,  produced. 

Proof  1.  1.  Because  A'B'  is  tangent  at  the  middle  point 
of  the  2ixc  ABi  it  is  parallel  to  the  chord  of  that  arc. 

2.  Because  the  two  polygons  are  equiangular  (Th.  IV.) 
and  have  a  pair  of  homologous  sides  parallel,  all  the  other 
homologous  sides  are  parallel.     Q.E.D. 

Proof  II.  Because  the  line  B'O  is  drawn  from  the  inter- 
section of  the  two  tangents,  B'U  and  B'T,  to  the  centre  of 
the  circle,  it  bisects  the  arc  TU. 

Therefore  it  passes  through  the  middle  point  B  of  this 
arc,  and  OBB'  are  in  the  same  straight  line  from  the  centre 
of  the  circle. 

457.  Oorollary.  The  polygons  may  also 
be  so  placed  that  the  circumscribed  polygon 
shall  touch  the  circle  at  the  angles  of  the 
inscribed  polygon. 

Theorem  VI. 

458.  The  greater  the  number  of  sides  of  a  regular 
circumscribed  polygon  theless  will    «^  .     .        f,,^, 
be  its  perimeter. 

Hypothesis.  BB',  one  side  of  a  reg- 
ular circumscribed  polygon  having  n 
sides;  00',  one  side  of  another  such 
polygon  having  n-\-l  sides;  OA,  a  per- 
■nendicular  from  tlie  centre  UT»on  BB\ 

Oonclusion.  The  perimeter  of  the 
polygon  having  n  sides  is  greater  than  that  having  w  -{- 1  sides. 


C0N8TRU0TI0N8. 


219 


Proof,    1.  Because  the  one  polygon  has  n  equal  sides,  and 
the  other  w  -f  1  equal  sides,  we  have 

Anglo  0"  0(7  =  -^. 

Angle  B'OB=  ^^^\ 
Therefore 

Angle  O'OG  :  angle  B'OB  ::n:n-{-l,  (§337) 

And  because  AOO  andAOB  are  respectively  the  halves  of 
C'OCand  B^OB, 

Angle  AOO  :  angle  A  OB  ::n  :n-{-l. 
That  is,  if  we  divide  the  angle  A  00  into  n  equal  parts,  A  OB 
will  be  %  + 1  of  these  parts.    Hence  GOB  will  be  one'of  the 
parts. 

2.  Divide  the  angle  AOC  into  n  parts  by  straight  lines 
meeting  AB  in  the  points  «,  b,  etc. 

Because  OA  is  a  perpendicular  upon  BB',  each  of  the  seg- 
ments J «,  ah,  10,  etc.,  will  be  longer  than  the  segment  next 
preceding  (§116).  Therefore  n  times  segment  OB  will  be 
greater  than  the  sum  of  the  n  segments  which  make  up  AG, 

3.  The  perimeter  of  the  polygon  of  n  sides  is 

nAB  =  nAG-{-nGB, 
and  that  of  the  polygon  of  w  +  1  sides  is 

{n-\-l)AG  =  nAG-\-AG, 

4.  Because  n6'^  is  greater  than  ^C, 

nAB>{n  +  l)AO; 
that  is,  the  perimeter  of  the  polygon  of  n  sides  is  the  greater. 

Q.E.D. 


•  ♦  > 


CHAPTER     II. 
CONSTRUCTION  OF  REGUUR  POLYGONS. 


^^^' J^]^^^^^^  i-  of  this  book  shows  that  a  regular  poly- 
gon of  anj,  uumber  of  sides  may  be  inscribed  in  a  circle  by 
dividing  the  latter  into  as  many  equal  parts  as  the  polygon 
has  sides,  and  joining  the  points  of  division  by  chords.    Hence 


220    BOOK  VI.    REQULAB  POLYGONS  AND  THE  GIROLE. 


the  problem  of  constructing  such  polygons  is  reduced  to  that 
of  dividing  a  circle  into  any  required  number  of  equal  parts. 

The  following  theorem  may  be  seen  almost  without  demon- 
stration. • 

If  we  can  divide  a  circle  into  any  number  of  equal  parts, 
we  can  also  divide  it  into  twice  that  number. 

For  if  we  divide  it  into  n  parts,  we  have  only  to  bisect  each 
of  these  parts,  which  we  do  by  §  271,  and  it  will  be  divided 
into  2n  parts. 

460.  It  is  also  easily  seen  that  the  problem  of  dividing 
an  arc  into  any  number  of  parts  may  be 
reduced  to  that  of  dividing  the  angle  cor- 
responding  to  the  arc  into  the  same  number 
of  parts. 

For  let  ABhk  the  arc,  and  0  its  centre. 

If  we  divide  the  arc  into  any  number  of 
equal  parts,  and  join  0  to  the  points  of 
division,  the  angle  A  OB  will   be  divided  into  that  same 
number  of  equal  parts. 

Hence,  by  dividing  the  arc  we  divide  the  angle, 

461.  Conversely,  if,  having  an  angle  A  OB,  we  draw  the 
arc  AB  of  a  circle  around  the  vertex  . 

0  as  a  centre,  and  divide  the  angle         ^,-\''       "?"^^^ 

A  OB  into  any  number  of  equal  parts 

by  the  lines  Oa  and  Ob,  meeting  the  Av, 

arc  in  a  and  b,  the  points  a  and  b 

will  divide  the  arc  into  that  same 

number  of  equal  parts. 

Hence,  by  dividing  the  angle  we  divide  the  arc. 

The  problem  of  bisecting  the  angle  (or  arc)  is  so  simple  that  it 
offered  no  difficulty  to  the  ancient  geometers.  By  bisecting  the  halves 
of  each  arc  it  might  be  divided  into  fourths,  and  so  on ;  therefore 
there  was  no  difficulty  in  dividmg  any  angle  or  arc  into  3,  4,  8,  16,  etc., 
equal  parts. 

This  being  so  easy,  it  was  naturally  sought  to  trisect  the  angle,  or 
divide  it  into  three  equal  parts. 

This  problem  of  the  tri  section  of  the  angle  was  long  celebrated. 
But  geometers  never  succeeded  in  sol"  "ng  it,  and  it  is  now  considered 
impossible  by  the  constructions  of  elementary  geometry. 


COmTRUCTIONS. 


221 


Problem  I. 

462.  To  dimde  a  given  circle  into  2,  4,  8, 16,  etc., 
equal  parts. 

Any  diameter,  as  AB,  will  divide  the  circle  into  two  equal 
parts  at  the  points  A  and  B. 

Drawing  through  the  centre  0  an- 
other diameter  perpendicular  to  this,  the 
circle  will  be  divided  into  four  equal  ^  I 
parts  by  four  radii,  at  right  angles  to 
each  other. 

By  joining  the  ends  of  these  radii  a 
square  will  be  described  in  the  circle. 

Bisecting  each  of  the  right  angles  at  the  centre  by  radii 
like  ON,  the  circle  will  be  divided  into  eight  equal  parts. 

By  joining  the  points  of  division,  a  regular  octagon  will  be 
inscribed  in  the  circle. 

The  process  of  bisection  may  be  continued  indefinitely,  so 
as  to  divide  the  circle  into  3"»  equal  parts,  where  m  may  be 
any  positive  integer. 


null 
id 


Problem  II. 

463.  To  divide  the  circle  into  3,  6,  12,  24,  etc., 
equal  parts. 

Analysis.   1.  Suppose  the  division  into  six  parts  effected, 
and  the  points  of  division  to  be  A,  B, 
C,  2>,  E,  F. 

3.  Draw  the  radii  OA,  OB,  etc.,  and 
join  AB,   BG,    CD,    etc.,   forming    Q,Ji 
regular  hexagon. 

3.  Because  each  of  the  three  equal 
angles  AOB,  BOO,  GOD,  is  one  third 
the  straight  angle  A  OD,  they  are  each 
angles  of  60°.     And  because  the  sides  OA,  OB,  are  equal,  the 

nnrrloa    f)  J  Ti  n^j/l    /ITf  4    <,r 


v----t-i--    i.cii\j.    \^SJJ1.    tlic   £ 


«1 1 


4.  But  the  sum  of  the  three  angles  is  180°.   Therefore  the 
three  angles  are  each  equal  to  00°;  that  is,  the  triangle  OAB 


EiU 


322    BOOK  VI.    REGULAR  POLYQOm  AND  THE  CIRCLE. 

is  equiangular  and  therefore  equilateral,  and  the  other  five 
triangles,  being  identically  equal  to  it,  are  also  equilateral. 

6.  Therefore  each  of  the  sides,  AB,  BG,  CD,  DE,  EF, 
Fa,  is  equal  to  the  radius  OA  of  the  circle.    Hence 

Construction.  1.  Starting  from  any  point  A  on  the  circle, 
cut  off  the  distances  AB,  BG,  GD,  etc.,  each  equal  to  the 
radius. 

2.  Six  equal  measures  will  T  to  the  point  A,  and  the 
circle  will  be  divided  into  six  ei^i  parts  M.  the  points  A,  B, 
(7,  etc. 

3.  The  alternate  points  A,  G,  E,  or  B,  D,  F,  divide  the 
circle  into  three  equal  parts. 

4.  By  bisecting  the  arcs  AB,  BG,  etc.,  the  circle  will  be 
divided  into  12  parts;  by  bisecting  these  arcs,  the  number  of 
parts  will  be  24,  etc. 

Corollary.  The  perimeter  AB -\- BG-\- GD-\- BE -\- EF 
+  FA  of  the  hexagon  ABGDEF  is  six  times  the  radius  of 
the  circle,  and  therefore  three  times  its  diameter. 

Because  each  of  the  six  sides  is  a  straight  line,  it  is  less 
than  the  corresponding  arc;  that  is,  side  AB  <  arc  AB,  etc. 
Therefore 

The  sum  of  the  six  arcs,  or  the  whole  circumference  of  the 
circle,  is  more  than  three  tirnes  the  diameter. 

Problem  III. 

464.  To  divide  a  circle  into  5,  10,  20,  etc.,  equal 
parts. 

Analysis.   Let  0  be  the  centre  of  the  circle,  and  the  arc 
AB  one  tenth  part  of  the  circumference, 
or  36°. 

Join  OA,  OB,  and  AB.    Then— 

1.   Because    the  sum  of  the  three 
angles  of  the  isosceles  triangle  A  OB  is  ^ 
180°,  the  sum  of  the  angles  A  and  B  7 
is  180°  -  36°  =  144°,  and  each  of  these  { 
angles  is  72°.     Therefore  each  of  the  \ 
angles   OAB  and  OB  A  is  double  the 
angle  at  0. 


OL' — • 


CONSTRUOTIONS. 


223 


2.  If  we  bisect  the  angle  OB  A  by  the  line  BPy  meeting 
OA  in  P,  the  angles  OBP  and  PBA  will  each  be  3G°,  and 
we  shall  have 

Angle  APB  =  180°  -  angle  PAB  -  angle  PBA       (8  74) 
=  180°  -  72°  -  36°  =  72°. 

3.  Therefore  the  triangle  BAP  is  isosceles  and  equiangu- 
lar, and  therefore  similar,  to  the  triangle  OAB.  Also,  because 
angle  POB  =  angle  PBO,  the  triangle  OPB  is  isosceles. 
Hence 

PO  =  PB  =  AB. 

4.  Because  of  the  similarity  of  the  triangles  BAP  and 
OAB, 

AP  :  AB  ::  AB  :  AO, 
or  ' 

AP  :  PO  ::  PO  :  AO, 

5.  Therefore  the  radius  OA  is  divided  in  extreme  and 
mean  ratio  at  the  point  P  (§  440).    Hence  we  conclude: 

465.  If  the  radius  of  a  circle  be  divided  in  extreme  and 
mean  ratio,  the  greater  segment  will  be  the  chord  of  one  tenth 
of  the  circle. 

Construction.  Divide  the  radius  OA  of  the  circle  in  ex- 
treme and  mean  ratio  at  the  point  P  (§  441). 

Around  ^  as  a  centre,  with  a  radius  equal  to  the  greater 
segment,  OP,  describe  a  circle,  and  let  B  and  Che  the  points 
in  which  it  intersects  the  first  circle. 

The  arc  AB  will  be  one  tenth  of  the  circle  and  i^C  will  be 
one  fifth  of  the  circle.  By  measuring  BCoE  five  times  around 
the  circle,  the  latter  will  be  divided  into  five  equal  parts. 

By  successive  bisection  the  circle  will  be  divided  into  10, 
20,  40,  etc.,  equal  parts.    Q.E.F. 


!!l 


l!ii 


1 


III  tin  III! 


U 


Problem  IV. 

466.  To  divide  a  circle  into  fifteen  equal  parts. 

Construction.  1.  From  any  point  A  cut  oft  AB  equal  to 
one  third  the  circle  (^  463V 

2.  From  the  same  point  A  cut  off  AD,  equal  to  one  fifth 
the  circle  (§  466). 


224    BOOK  YL    REGULAR  POLYGONS  AND  THE  CIRCLE. 


3.  Arc  BD  will  then  be  (^  —  \)  of  the  circle;  that  is,  ^ 
of  it. 

4.  Therefore,  if  we  bisect  the  arc 
BD,  we  shall  have  an  arc  equal  to  ^ 
of  the  circle.  By  measuring  this  arc 
off  16  times,  the  circle  will  be  divided 
into  15  equal  parts.    Q.E.F. 

Scholium.  The  foregoing  divi- 
sions of  the  circle  are  all  that  were 

known  to  the  ancient  geometers.  But,  a 

about  the  beginning  of  the  present  century.  Gauss,  the  great 
mathematician  of  Germany,  showed  that  whenever  any  power 
of  3,  increased  by  1,  made  a  prime  number,  the  circle  could 
be  divided  into  that  number  of  parts  by  the  rule  and  com- 
pass.   Thus: 

2^'-|-  1  =     3,  a  prime  number. 

2»  +  1  =     5 

2*  +  l=    17 

2"  +  1  =  257 

Therefore,  besides  the  old  solutions,  the  circle  can  be 
divided  into  17  or  into  257  equal  parts. 

The  division  into  17  parts  by  construction  is,  however,  too 
complicated  for « the  present  work,  and  that  into  257  parts  is 
so  long  that  no  one  has  ever  attempted  to  really  execute  the 
construction. 


it 


<t 


i( 


« 


tt 


<t 


•  * 


CHAPTER    III. 

AREAS   AND    PERIMETERS    OF    REGULAR    POLYGONS 

AND  THE  CIRCLE. 


46*7.  Def.  The  apothegm  of  a  regular  polygon  is 
the  perpendicular  ironi  its  centre  upon  any  one  of  its 
sides. 


AREAS. 


225 


B 


Theobem  VII. 

468.  The  area  of  a  regular  polygon  is  equal  to 
half  the  rectangle  contained  by  Us  p&rirmter  and 
its  apothegm,  b  u 

Hypothesis.    ABCDEF,  a  regular 
polygon;  OP,  its  apothegm. 

Conclusion. 
Area  ABCDEF  =  ^OP  x  perimeter.  ^^  ^' '^^ 

Proof.  Join  OA,  OB,  etc.    Then— 

1.  Because  OP  is  the  altitude  of 
the  triangle  A  OB,  ^       P 

ATesiAOB  =  ^OP.AB. 

2.  In  the  same  way  it  may  be  shown  that  the  area  of  each 
of  the  other  triangles  into  which  the  polygon  is  divided  is 
equal  to  one  half  of  the  side  into  the  apothegm.  But  the 
apothegms  are  all  equal.     Therefore 

Area  ABCDEF =iOP.  AB  +  ^OP.  BC-\-^OP.  Ci>  +  etc. 
=  iOP  X  perimeter.  Q.E.D. 

469.  Corollary  1.  Because  the  perimeter  of  each  cir- 
cumscribed regular  polygon  is  less  the  greater  the  number  of 
its  sides  (§458),  it  follows  that  the  area  of  the 
circumscribed  regular  polygon  is  1p«?s  the  greater 
the  number  of  its  sides. 

Cor.  2.  It  is  easily  shown  that  the  area  of 
the  circumscribed  square  is  equal  to  the  square 
upon  the  diameter  of  the  circle.     Therefore: 

470,  The  area  of  any  circumscribed  regular  polygon  of 
more  than  four  sides  is  less  than  the  square  upon  the  diameter 
of  the  circle. 

4*71.  Scholium.  If  a  regular  polygon  be  inscribed  in  a 
circle,  and  another  regular  polygon  of  the  same  number  of 
sides  be  circipiscribed  about  it,  the  area  of  the  outer  poly- 
gon will  be  greater  than  that  of  the  inner  one  by  the  surface 
contained  between  the  perimeters  of  the  two  polygons. 
This  surface  will  be  called  the  included  area."  ° 
When  the  two  polygons  are  so  pla<5ed  that  their  respectivo 


j     il 


III  I  li;f 


226   BOOK  VI.   BEQULAB  POLYGONS  AND  THE  CIRCLE. 

sides  are  parallel  (§  466),  the  included  area  will  be  formed  of 
as  many  identically  equal  trapezoids,  AA'B'By  BB'G*Cy  etc., 
as  the  polygons  each  have  sides;  and  each  apothegm,  as  OT, 
will  again  divide  each  of  these  trapezoids  into  two  iden- 
tically equal  trapezoids. 

When  rhe  polygons  are  so  placed  that 
the  sides  of  the  circumscribed  polygon 
shall  touch  the  circle  at  the  vertices  of  the 
inscribed  polygon  (§457),  the  included  area 
is  made  up  of  as  many  identically  equal 
triangles  as  the  polygons  have  sides. 


Theorem  VIII. 

473.  WTieT^  the  inscribed  and  circumscribed  regu- 
lar polygons  ham  more  than  four  sides,  the  included 
area  is  less  than  the  square  upon  one  side  of  the 
inscribed  polygon. 

Hypothesis.    0,  the  centre  of  the  circle;  BG,  a  side  of  the 
inscribed  polygon;  FG,  a  side  of  xhe 
circumscribed  polygon,  placed  paral- 
lel to  BC. 

Conclusion.  If  the  polygons  be 
completed,  the  included  area  will  be  ^i 
less  than  the  square  upon  CB. 

Proof.  Join  OBF.  Drop  the 
perpendicular  OQ  from  0  upon  FG; 
draw  the  diameter  COE  and  join 
BR.    Let  n  be  the  number  of  sides  of  each  polygon.    Then — 

1.  Because  tl  3  area  of  the  inscribed  polygon  is  made  up 
of  2n  triangles  identically  equal  to  OBF,  and  the  circum- 
scribed polygon  of  2n  triangles  equal  to  OFQ,  the  inscribed 
polygon  will  be  to  the  circumscribed  one  as  the  area  OBP 
is  to  the  area  OFQ.    That  is,  if  we  put 

A,  the  area  of  the  circumscribed  polygon, 
a,  the  area  of  the  inscribed  polygon, 
we  shall  have 

A  :a  :: area  OFQ  :  arcE  OBP, 


AREAS. 


227 


Hence,  by 
(§365) 


2.  Because  the  linos  BO  and  FO  are  parallel,  the  tri- 
angles OJJP  and  OFQ  are  similar.    Because  of  this  similarity. 

Area  OFQ  :  area  OBF  ::  OQ'  :  0P\         (§423) 

3.  Comparing  with  (1),  and  because  OQ  =  OB,  both  being 
radii  of  the  same  circle, 

Aia  I'.OB^ :  0P\ 

4.  The  included  area    is   equal  to  A  —  a. 
division, 

A-a:A:'.0B^-  OP^ :  OB*, 
6.  Because  OFB  is  a  right-angled  triangle, 

OB'  -  OP'  =  PB\ 
Making  this  substitution  in  (4),  and  putting 
A  the  diameter  of  the  circle  (whence  OB  =  ^D), 
s,  the  length  of  the  side  CB  of  the  inscribed  polygon  (whenca 

PB  =  ^8), 
we  shall  have 

A-a:A::is':iD'::s':D\  (346) 

Therefore 

8*  X  A  _  A^ 


A  —  a  = 


B'      ~  Z) 

6.  But  A  <  D\  or  ^  -.  i)»  <  1  (§  470).    Therefore 
A-a<8\     Q.E.D. 

Corollary,    %  sufficiently  increasing  the  number  of  sides 
of  the  polygons,  we  can  make  each  side  as  short  as  we  please 
and  therefore  its  square  as  small  as  we  please.     Hence:  ' 

473.  If  the  number  of  sides  of  the  inscribed  and  cir- 
cumscribed polygons  be  indefinitely  increased,  the  included 
area  will  become  less  than  any  assignable  quantity. 

Problem  V. 

474.  From  the  areas  of  the  inscribed  and  circum- 
scribed polygons  of  n  sides  to  find  the  areas  of  those 
hamng  2n  sides. 

Given.  0,  the  centre  of  the  circle;  AB.  one  of  t.h«  «i^os 
Of  the  inscribed  polygon  of  n  sides;  OD,  one  of  the  sides  of 
tHe  circumscribed  polygon  of  n  sides,  placed  paraUel  to  AB 


Hill 


fl 


Jill 


t\ 


228   ^OOK  VI.   UEQULAli  rOLYUOm  AI/JJ  X^^  CUWLE. 


and  tangent  to  the  circle  at  F\  OEF,  the  perpendicular  from 
the  centre  upon  the  tungeut;  AF^  FJJ,  two  sides  of  the  in- 
scribed polygon  of  2n  sides;  A  0, 
BH,  tangents  to  the  circle  at  A 
and  B;  wherefore  OH  is  one  side 
of  the  circumscribed  polygon  of 
2n  sides  (though  not  parallel  to 
any  side  of  the  inscribed  polygon 
of  2n  sides).  Also,  the  areas  of 
the  triangles  OAB  and  OCB  are 
supposed  given,  and  from  them 
those  of  the  inscribed  and  circumscribed  polygons  are  found 
by  multiplying  by  n. 

Required.  I.  To  find  the  area  OAF  (from  which  the 
area  of  the  second  inscribed  polygon  is  obtained  by  multiply- 
ing by  3n).         ' 

II.  To  find  the  area  OOH  (from  which  the  area  of  the 
second  circumscribed  polygon  is  found  by  multiplying  by  2w). 

Solution.     Let  us  put 
ty  the  area  of  the  triangle  OEA,  which  is  one  half  that  of  the 

given  area  OAB; 
T,  the  area  of  the  triangle  OFC,  which  is  one  hdf  that  of 

the  given  area  OCB; 
t\  the  required  area  OAF; 
T',  the  required  area  OGH.    Then — 

1.  1.  Because  the  triangles  OAF  skud  O^i^have  the  same 
vertex  A,  and  their  bases  on  the  same  straight  line  OF,  their 
areas  are  as  OF  to  OF,  or 

t:t'  ::  OF:  OF.  (§416) 

2.  Because  the  triangles  OAF  and  0 OF  have  the  common 
vertex  F,  and  their  bases  on  the  same  straight  line  OC, 

t' :  T '.:  OA  :  OG. 

3.  Because  ^^and  GF  are  parallel, 

OF :  OF ::  OA  :  OG. 

4.  Comparing  with  (1)  and  (2), 

t:t'  ::f  :  T; 
that  is,  the  area  t'  is  a  mean  proportion  between  t  and  T.  or 

V=  VIT, 


AJiBAJS. 


229 


GF 


II.  6.  Because  the  triangloa  FOO  and  A  GOhcLYi 
OA  =  OF,  and  00  coinnion,  thoy  arc  identically  o<iual.    Also 
because  OGA  and  OGC  have  the  same  vertex  G,  thoy  are  to 
each  other  as  OA  to  00.     Hence 

Area  FOG  :  area  GOG ::  OJ  :  00 ::  /' :  T.  (2) 

0.  But  ^  ^ 

Area  FOG  =  |  area  GOir=  \T'. 

Area  GOO—  area  FOO -  FOG  =  T—yr\ 

7.  Comparing  with  (5), 

iT':T~iy"::/':r; 
and  by  composition, 

whence  7"  = 

8.  If  wo  put  yl,  ^',  the  areas  of  the  circumscribed  poly- 
gons; a,  a',  the  areas  of  the  inscribed  polygons,  wo  shall  have 

A  =  2nT,         a    =  2wj?, 
A'  =  2nT',       a'  =  2nt', 
and  the  relations  between  A,  A',  a,  and  a'  will  bo  the  same  as 
those  between  T,  T\  t,  and  t'  in  (4)  and  (7). 
We  therefore  conclude: 

If  we  have  given  the  areas  A  and  a  of  the  circumscribed 
and  inscribed  polygons  of  n  sides,  those  of  the  corresponding 
polygons  of  2n  sides  will  be  given  by  the  equations 

fl'=  Vol, 
.,  _    2a' A 

a'  +  A' 

4*7  6.  Application  of  the  preceditig  solution  to  the  compu- 
tation of  the  area  of  inscribed  and  cir- 
cumscribed polygons  of  a  continually 
increasing  number  of  sides. 

Let  us  put 
r,  the  radius  of  the  circle. 

I.  Let  us  begin  with  the  inscribed 
and  circumscribed  regular  hexagons. 

Each  of  these  hexagons  can  be 
uiyided  into  six  equal  equilateral  tri-  - 

angles  by  lines  drawn  from  the  centre  (§  463).    Each  side  of 


111' 


»Hi 


r 


\\- 


230   BOOK  VI.    REGULAR  POLYGONS  AND  THE  CIRCLE. 

the  triangles  in  the  inscribed  hexagon  will  be  equal  to  r. 
The  area  of  each  equilateral  triangle,  found  by  the  method  of 
§333,  will  be 

and  the  area  of  the  whole  inscribed  hexagon  will  be 


6  4^    .      SV'S 


r  = 


r\ 


4  2 

To  find  the  area  of  the  circumscribed  hexagon  we  have 
OM  :  OB' ',:  ON '.  OB, 
or  OM  ir  y.r  :  OB, 

whence  for  the  side  of  the  circumscribed  hexagon 


0B  = 


r' 


2r 


Because  OAB  is  an  equilateral  triangle. 


Area  OAB  = 


Vs 


0B'=: 


Vd  4 


_  f 
4    3 


V3" 


whence  for  the  area  of  the  whole  circumscribed  hexagon 

-4rr  r"  =  2  V3r\ 

Then,  using  the  previous  notation,  we  have 


a  = 


2 


r . 


A  =2  4^r'. 

II.  Polygon  of  12  sides.     Next  we  pass  to  the  polygons  of 
12  sides  by  the  formulaB 

a'=  V^;  A'  =  ^.. 

a'  -{-A 

Making  these  substitutions,  and  reducing  by  algebraic 
methods,  we  have 


A'  = 


12 


2-\-  V3 


^  r'  =  3.2153910r' 


ABBA8. 


S31 


This  value  of  o'  gives  the  remai-kable  result  that  the  area 
S™™^'""'^  '*^''^  P^^ye""  °*  12  sides  is  exact  ytCe 

12  »m!;  fif  ?/  ^*  '"■''^-    To  P»^  from  the  polygon  of 

If  we  put  «- and  ^or  the  areas  required,  we  have 

.A-7-7--  __  V3  X  3.21539ir', 


fl"  = 


^"  = 


2a'M' 


or,  reducing  to  numbers,  ^    ^ 

«"  =  3.105829r', 
^"  =  3.159661r». 

96  sides,  and  so  on  indefinitely.  ' 

Subtracting  the  area  of  each  inscribed  one  from  fyi«+  .f 
the  coiTesponding  circumscribed  one  we  have  th^n^il  f 
area  in  each  case.     The  results  fn  iqo    /  ,     included 

following  table.  ^^^  ''^''  ^  '^^^^  ^^  *lie 

Area  of 

circumscribed 

polygon. 

3.215391r» 
3.159661r» 
3.146087r' 
3.142715r' 
3.141873r« 


No.  of 
Bides. 


a. 

Area  of 
inscribed 
polygon. 

13 3.000000r'' 

24....  3.105829/-' 

48....  3.132630r» 

96....  3.139350r' 

192....  3.141032r» 


A  — a. 

Included 

area. 


0.215391/-' 
0.053832P 
0.013457r* 
0.003365r» 
0.000841r» 


(§473Vthough  we  can  never  redlltto"^.'"  ""'  *^*^"* 

Area  of  the  Circle. 

fixe?a;a^l  '^^,„^*  °f  ^  -^f-g  magnitude  is  a 

approach  "so'^s"  to  a^XrZHSir:*^^'.^^^  "^^ 
please,  but  to  which  it  c.n  „e"e?£r;„T  "  "^ 


sil 


u 


HI 


i 


232  BOOK  VI.    REGULAR  POLYGONS  AND  THE  CIRCLE. 

477.  Theorems  of  Limits.  The  theorems  relating  to  the 
subject  of  limits  are  proved  in  algebra.  The  following  propo- 
sitions are  applied  in  geometry: 

Axiom  I.  Any  quantity  may  he  multiplied  ly  a  factor  so 
great  as  to  inake  the  product  exceed  any  quantity  we  may  assign. 

Ax.  II.  Any  quantity  may  le  multiplied  by  a  factor  so 
small  as  to  make  the  product  less  than  aiiy  quantity  we  may 
assign. 

Theorem.  If  two  varying  quantities  each  ap- 
pToacli  a  limit,  the  limit  of  their  product  will  be  the 
product  of  their  limits. 

Hypothesis.     A  quantity  X  approaching  the  limit  L, 
A  quantity  X'  approaching  the  limit  L', 

Conclusion.  The  product  XX'  will  approach  the  product 
LL'  as  its  limiti 

Proof.     Let  a  and  a'  be  the  respective  amounts  by  which 
-Tand  X'  differ  from  their  limits  L  and  L\    We  then  have 
X  •=■  L  —ay 

X=  L'  -a\ 

Multiplying, 

XX  =  LU  -  aU  -  a'L  -f  aa\ 

Let  /?  be  a  quantity  as  small  as  we  please.  How  small 
soever  it  may  be,  we  may  take  the  quantities  a  and  «'  so  small 
that  the  products  aU,  a'L,  and  aa'  shall  each  be  less  than 
one  third  of  /?.     (Ax.  II.) 

The  quantity  XX'  will  then  differ  from  LU  by  less  than  /?. 

Because  the  difference  /?  may  be  as  small  as  we  please,  the 
product  LL'  is  the  limit  of  XX.    Q.E.D. 

Gor.  Because  the  area  of  a  rectangle  is  represented  by 
the  product  of  the  lengths  of  its  containing  sides  we  conclude: 

If  the  containing  sides  of  a  rectangle  approach  two  lines  L 
and  L'  as  their  limits,  the  area  of  the  rectangle  will  approach 
the  area  LL'  as  its  limit. 

Lemma. 

478.  When  the  number  of  sides  of  the  inscribed 

each  of  their  areas  approaches  the  area  of  the  circle 
as  its  limit. 


ABBAS. 


233 


Proof.    In  order  to  prove  this  lemma  we  have  to  show: 
I.  That  the  area  of  the  inscribed  polygon  must  always 
be  less  than  that  of  the  circle,  how  great  soever  the  number  of 
its  sides. 

II.  That  the  area  of  the  circumscribed  polygon   must 
always  be  greater  than  that  of  the  circle. 

III.  That  if  we  assume  an  area  a,  we  can  by  increasing 
the  number  of  sides  of  the  polygons  make  each  of  their  areas 
differ  from  that  of  the  circle  by  less  than  a,  how  small  soever 
a  may  be. 

1.  Because  the  apothegm  OJf  (§  475)  of  the  inscribed  regu- 
lar  polygon  is  a  perpendicular  from  0  upon  A'B',  it  is  less 
than  the  line  OA',  which  is  the  radius  of  the  circle. 

Therefore  some  part  of  the  area  of  the  circle  will  always 
be  outside  the  polygon,  and  the  area  of  the  polygon  must 
always  be  less  than  that  of  the  circle. 

2.  In  a  similar  way  we  may  show  that  the  area  of  the  cir- 
cumscribed polygon  is  always  greater  than  that  of  the  circle. 

3.  How  small  soever  the  area  a,  we  may  increase  the  num- 
ber of  sides  of  the  polygons  until  the  included  area  shall  be 
less  than  a  (§  473). 

Then,  because  the  area  of  the  circle  is  greater  than  that  of 
the  inscribed  polygon,  but  less  than  that  of  the  circumscribed 
polygon,  it  will  differ  from  each  of  them  by  less  than  a. 

Therefore  the  area  of  the  circle  is  a  quantity  which  the 
area  of  each  polygon  may  approach,  so  as  to  differ  from  it  by 
less  than  any  assignable  quantity,  but  to  which  it  can  never 
become  equal. 

Therefore  the  area  of  the  circle  is  the  limit  of  the  area  of 
each  polygon  when  the  number  of  its  sides  is  indefinitely  in- 
creased.    Q.E.D. 

4*79.  By  continuing  the  table  of  §  475  we  may  approxi- 
mate as  nearly  as  we  please  to  the  area  of  the  circle.  But  there 
is  a  theorem  of  approximation  which  we  give  without  proof 
and  which  will  enable  us  to  make  a  more  rapid  approximation! 
i'  We  oUain  an  approximate  area  of  the  circle  hy  adding 
to  the  inscribed  poly f/G, I  two  thirds  of  the  included  area. 

II.   Tliis  approximation  is  nearer  the  truth  the  greater  the 
number  of  sides. 


iHli:i 


234   BOOK  VI.    REGULAR  POLYGONS  AND  THE  CIRCLE. 

Applying  this  rule  to  the  preceding  table,  we  have- 
No  of  Bides.  K^  -a).  o + «4  -  a). 

12 •143694r*  3  •143594r' 

34 •035888r"  3  •1417l7r« 

48 -OOSQTlr'  3-141601r' 

96 •002243r'  3'141693r' 

193 -OOOSGlr'  3-141593r« 

We  see  that  we  get  the  same  result  from  96  sides  and  192 
sides,  so  that  they  both  may  be  regarded  as  correct  to  the  sixth 
place  of  decimals. 

480.  The  coefficient  of  r\  which  we  have  found  to  six 
places  of  decimals,  is  represented  by  the  symbol  n.    That  is 
we  put  ' 

n  =  3.141593  .... 
nr"  =  area  of  circle  of  radius  r. 

481.  Corollary  TJie  areas  of  any  two  circles  are  pro- 
portional to  the  squares  of  their  diameters. 

Circumference  of  the  Circle. 

483.  Axiom.  When  we  increase  indefinitely  the 
number  of  sides  of  the  inscribed  and  circumscribed 
polygons,  the  perimeter  of  each  of  these  polygons 
approaches  the  circumference  of  the  circle  as  its  limit. 

Theorem  IX. 

483.  The  area  of  a  circle  is  equal  to  one  half  «^« 
radius  into  its  circumference. 

Proof  1.  Let  a  regular  polygon  of  n  sides  be  circum- 
scribed about  the  circle. 

The  area  oi  this  polygon  is  equal  to  half  its  apothegm  into 
its  perimeter,  and  its  apothegm  is  equal  to  the  radius  of  the 
circle. 

Let  the  number  of  sides  of  the  polygon  be  increased 
indefinitely.     Then — 

2.  The  area  of  the  polygon  will  approach  the  area  of  the 
circle  as  its  limit. 

3.  The  perimeter  of  the  polygon  will  approach  the  circum- 
ference of  the  circle  as  its  hmit  (§  482). 


CIRGUMFERBNOB  OF  THE  GIRGLE. 


235 


7t', 


4.  Therefore  the  limit  of  area  (area  of  the  circle)  will  be 
equal  to  half  the  radius  into  the  limit  of  the  perimeter  (cir- 
cumference of  the  circle)  (§  478).     Q.E.D. 

Problem  VII. 

484.  To  find  the  ratio  of  the  circumference  of  the 
circle  to  its  diameter. 

Put  C\  the  circumference;  Z>,  the  diameter,  or  2r;  A^  the 
area  of  the  circle. 
By  §  480  we  have 

A^nr"  =  \7tD\ 
But  we  have  just  proved  that 

A=^rC=\DG, 
Therefore  \DG  =  InD^; 

or  0=ytD, 

D 

that  is,  the  number  it  =  3.14159 

the  circumference  of  the  circle  to  its  diameter. 

This  number  ic  is  of  such  fundamental  importance  in  geometry  that 
mathematicians  have  devoted  great  attention  to  its  calculation.  The 
preceding  method,  by  which  we  have  found  it  to  six  decimals,  is  the 
easiest  afforded  by  elementary  geometry,  but  more  rapid  methods  are 
afforded  by  the  higher  mathematics.  Dasb,  a  German  omputer,  car- 
ried the  calculation  to  200  places  of  decimals.  The  followmg  are  the 
first  36  figures  of  his  result:* 

3.141  692  653  689  793  238  462  643  383  279  602  884. 
The  result  is  here  carried  far  beyond  all  the  wants  of  mathematics. 
Ten  decimals  are  sufficient  to  give  the  circumference  of  the  earth  to  the 
fraction  of  an  inch,  and  thirty  decimals  would  give  the  circumference 
of  the  whole  visible  imiverse  to  a  quantity  imperceptible  with  the  most 
powerful  microscope. 

EXERCISES. 

1.  Assuming  the  radius  of  a  circle  to  be  5  metres,  com- 
pute, by  the  process  of  §  475,  the  area  of  the  inscribed  and 
circumscribed  regular  hexagon,  dodecagon,  and  polygon  of 
24  sides. 


.  is  itself  the  ratio  of 


Crelle's  Journal,  Vol.  27,  p.  198. 


236    JiOOK  VI.    BEQULAR  P0LYO0N8  AND  THE  OIROLE. 

Note.  In  computations  like  this,  the  student  should  not  be  satisfied 
by  working  blindly  with  the  formula),  but  should  reason  the  results  out 
by  the  same  process  employed  to  reason  out  the  formula}.  In  the  present 
case  the  computation  of  the  area  of  the  hexagon  is  easy;  and  that  of  the 
figures  of  12  and  24  sides  can  then  be  executed  as  in  §  476. 


2.  If  an  equilateral  triangle  bo  inscribed  in  a  circle,  show 
that  the  perpendicular  from  any  vertex  upon  the  opposite 
side  is  three  fourths  the  diameter  of  the  circle. 

3.  Using  the  preceding  theorem,  compute  the  length  of 
sides  and  the  area  of  the  equilateral  triangle  inscribed  in  the 
circle  whose  radius  is  unity. 

4.  Show  that  tiie  altitude  of  a  circumscribed  equilateral 
triangle  is  three  times  the  radius  of  the  circle. 

5.  Without  using  any  of  the  preceding  theorems,  show 
that  the  radius  pf  the  circle  circumscribed  about  an  equilateral 
triangle  is  double  the  radius  of  the  inscribed  circle. 

6.  What  conclusion  thence  follows  respecting  the  relation 
of  the  areas  of  the  two  circles?    (§  481.) 

7.  If  the  radius  of  a  circle  is  r,  what  is  the  length  of  each 
side  of  the  circumscribed  equilateral  triangle  ? 

8.  In  a  circle  of  radius  r,  find  the  sides  of  the  inscribed 
and  circumscribed  squares  and  their  areas. 

9.  From  the  results  of  the  preceding  example  find  the 
areas  of  the  inscribed  and  circumscribed  regular  polygons  of 
8,  16,  33,  and  64  sides,  and  thence  the  area  of  the  circle,  as 
in  §§  475,  479. 

10.  A  bought  a  piece  of  pasturage  30  yards  X  40  yards  in 
B's  field,  and  then  tied  his  cow  in  the  centre  with  a  rope  just 
long  enough  to  reach  to  the  corners  of  his  piece.  Over  how 
much  of  B's  part  of  the  field  could  A's  cow  feed? 

11.  Four  equal  circles  of  radius  a  have  their  centres  on 
the  corners  of  a  square,  and  touch  each  other.  What  is  the 
radius  of  the  circle  in  the  centre  touching  each  of  them? 

12.  What  must  be  the  diameter  of  a  circle  in  order  that  its 
area  may  be  100  square  feet?    (Apply  §  480. ) 

13.  In  a  regular  polygon  of  n  sides,  what  angle  (in  degrees) 
floes  a  xine  froni  any  vertex  to  the  ccntro  make  with  the  sides 
meeting  at  that  vertex?    (§  160.) 


MAXIMUM  FIOUUES. 


237 


14.  If  from  one  vertex  of  a  regular  polygon  of  n  sides 
lines  be  drawn  to  all  the  other  vertices,  what  angles  will  they 
form  with  each  other?    (Apply  §  235.) 

16.  What  is  the  area  of  a  circle  circumscribed  about  a 
square  whose  side  is  a? 

16.  If  the  apothegm  of  a  regular  hexagon  is  h,  what  is  the 
area  of  the  ring  included  between  its  inscribed  and  circum- 
scribed circles? 


■• »  » 


CHAPTER    IV. 

MAXIMUM    AND   MINIMUM    FIGURES. 


485.   Def.   A  maximum 
figure  of  a  given  class. 


figure  is  the  greatest 


^  486.    Def.   A  minimum  figure  is  the  least  of  a 
given  class. 

Eemark.  If  a  figure  is  entirely  unrestricted,  there  can  be 
no  such  thing  as  a  maximum  or  a  minimum,  because  a  figure, 
if  not  restricted,  can  be  made  as  great  or  as  small  as  we  please! 

Hence  a  maximum  or  minimum  figure  means  one  subject 
to  certain  conditions;  for  example,  required  to  have  a  certain 
perimeter,  or  to  be  included  between  certain  limits,  or  to  have 
some  relations  among  its  parts  which  prevent  it  from  becom- 
ing indefinitely  great  or  indefinitely  small. 

Having  defined  the  conditions  of  the  figure,  we  may 
imagine  ourselves  to  construct  all  possible  figures  fulfilling 
these  conditions.  This  collection  of  possible  figures  will  con- 
stitute a  class.  The  greatest  among  them  will  be  the  maxi- 
mum;  the  least,  the  minimum.  * 

48T.  Def.  Isoperimetrical  figures  are  those  which 
have  the  same  perimeter. 


!„ 


238  BOOK  VI.    BEQULAR  POLYGONS  AND  THE  CIRCLE. 


Theobem  X. 

488.  If  two  sides  of  a  triangle  he  given,  its  area 
will  he  a  maximum  when  these  sides  are  at  right 
angles. 

Proof.  Let  AB  and  AP  be  the  two  given  sides  of  the 
triangle. 

At  whatever  angle  we  fix  these  |t^     p» 

sides  the    area  will    be  equal  to  y 
AB  X  altitude,  and  so  will  be  the  rv;--,, 
greatest    when    the    altitude   is  j    ^'\    / 
greatest  (§  301).  -^ -'-J- 

If  JP  is  perpendicular  to  ^i5, 
AP  will  itself  be  the  altitude.  In  any  other  position,  as 
AP'  or  AP",  thp  altitude  A'P'  or  A"P"  will  be  less  than 
^P(§101). 

Therefore  the  triangle  of  greatest  area  is  BAP,  in  which 
AP  L  AB.     Q.E.D. 

489.  Problem  VIII.     Having  given 

A  straight  line  M, 

Two  points  E  and  F on  the  same  side  of  the  line: 

It  is  required  to  find  the  point  P  on  the  line  M  for  which 
the  sum  of  the  distances  PE -^  PF  shall  le  a  minimum. 

Solution.    From  one  of  the  given  points,  as  F,  drop  a 
perpendicular  upon  the  line  M,  and 
produce  it  to  the  point  F'  at  an 
equal  distance  on  the  other  side. 

Because  the  line  M  is  the  per- 
pendicular bisector  of  FF',  every 
point  upon  it  will  be  equally  dis- 
tant from  F  and  F*  (§  104). 

Therefore,  if  P'  be  any  point  upon  the  line,  we  shall  have 
EP'  -f  P'P  =  EP'  +  P'i^. 

The  distance  EP'  +  P'P'  will  be  a  minimum  when  P'  is 
in  the  straight  line  from  E  to  F'.  Therefore  the  required 
point  P  is  the  point  in  which  the  straight  line  from  Eio  F' 
intersects  the  given  line. 

Draw  PF. " 


MAXIMUM  FI0URE8. 


239 


Because  the  line  M  is  the  perpendicular  bisector  of  the 
line  FF*,  we  haye 

Angle  MPF'  =  angle  MPF; 
also.  Angle  MPF'  =  opp.  angle  £PP; 

whence  Anglo  BPP'  =  angle  MPF. 

The  solution  of  the  problem  is  therefore  expressed  in  the 
following  lemma: 

490.  Lemma.  The  sum  of  the  distances  from  a  movable 
point  on  a  straight  line  to  two  fixed  points  on  the  same  side 
of  the  line  is  a  minimum  when  those  distances  make  equal 
angles  with  the  straight  line. 

Scholium.  If  the  line  JIf  is  a  section  of  a  mirror,  the 
lines  BP  +  PF  are  those  which  would  be  followed  by  a  ray 
of  light  emanating  from  a  candle  at  F  and  reflected  to  F, 
because  it  is  a  law  of  optics  that  the  angles  of  incidence  and 
reflection,  namely  FPP'  and  FPM,  are  equal.    Hence: 

The  course  taken  ly  a  ray  of  light  emanating  from  one 
point  and  reflected  hy  a  plane  surface  to  another  point  is  the 
shortest  path  from  the  one  point  to  the  reflector,  and  thence  to 
the  other  point. 


Theorem  XL 

491.  If  the  hase  of  a  triangle  and  the  sum  of  the 
other  two  sides  he  given,  the  area  will  he  a  maximum 
when  these  sides  are  equal. 

Hypothesis.  APB,  an  isosceles  tri- 
angle on  the  base  AB;  AP'B,  another 
triangle  on  the  same  base  AB,  in  which 

AP'-\-P'B  =  AP  +  PB. 

Conclusion. 

Area.  AP'B  <  area^P^. 

Proof.  Through  P  draw  PJ^IMJ?.     ^ 
Because  the  areas  of  the  triangles  are  ^ 
proportional  to  their  altitudes,  it  is  sufficient  to  show  that  the 
vertex  P'  must  fall  below  the  parallel  PF, 

1.  Because  angle  PAB  =^angle  PBA  (§  91),  and  PF  || 
AB,  the  sides  AP  and  PB  make  equal  angles  with  PF. 


i  A 


^40  BOOK  VI.    UKQULAIi  POLYGONS  AND  THE  CIRCLE. 


P'  cannot  lie  on  FF,  because  then  wo 


2.  The  vertex 
should  have 

AF'  H-  F'B  >AF-\.  FB.  (§  490) 

3.  If  possible,  suppose  the  vertex  F'  to  fall  at  any  point 
R  above  FE.  The  sides  RA  and  RB  will  then  include  a 
segment  of  the  line  FP  between  them. 
From  any  point  Q  of  this  segment 
draw  QA  and  QB.    Then 

AR+RB>AQ  +  BQ.     (§100) 
AQ -{- QB  >  AF -{.  FB,     (§490) 
Therefore 

AR+RB>  AF-\-FB. 
Because  R  may  be  any  point  above  ^ 
FF,  the  vertex  F'  cannot  fall  above  the  line  FF. 

4.  Since  it  can  fall  neither  above  nor  upon  this  lino,  it 
must  fall  below  it,  and  we  must  have 

Alt.  of  P'  <  alt.  of  F; 

Area  AF'B  <  area  AFB.     Q.E.D. 


whence 


Theorem  XII. 
492.  Among  all  isoperimetrical  polygons  of  a 
given  number  of  sides,  that  of  maximum  area  has  all 
its  sides  equal. 

Froof,  If  possible,  let  ABODEFhe 
the  maximum  polygon  of  given  perimeter 
and  number  of  sides  in  which  some  two  ^ 
adjacent  sides,  as  AB  and  BC,  are  un- 
equal. Join  AOy  and  describe  on  ^ Can 
isosceles  triangle  AB'Oy  such  that  *. 

AB'-\.B'0=AB  +  Ba     Then—  ^ 

1.  Because    ABO   is   isosceles    and 
AB''i-B'C=AB-{-BO, 

Area  AB'C  >  area  ABC. 

2.  Because  the  area  ACBBFremamB  unchanged. 

Area  AB'CDEF>  area  ABCDEF. 

3.  But  the  polygon  AB'GDEF  has  the  same  perimeter 
and  number  of  sides  as  the  polygon  ABCDEF.  Because  the 
xoi-iner  has  a  gTeater  area,  ^q  latter  cannot  be  the  polygon  of 
maximum  area. 


(§491) 


II 


MAXIMUM  FIOUHIBS. 


241 


4.  Therefore  no  polygon  having  two  adjacent  sides  unequal 
can  be  a  polygon  of  maximum  area;  and  because  any  polygon 
with  unequal  sides  must  have  some  two  adjacent  sides  imequal, 

rid^S'TEtE^'''"" "' "'"'""""  ^'^^  "--^  ^»-  ^  '^ 

Theorem  XIII. 

•493.  7)r  a  line  of  gimn  length,  which  may  he 
mrved  at  pleasure,  is  required  to  7tave  its  extremities 
upon  an  indefinite  straight  line,  it  will  inclose  a 
maximum  area  when  hmt  into  a  semicircle. 

Hypothesis.  MN,  an  indefinite  straight  line:  ADB  a 
curve  line  wliieh  may  bo  bent  at  ' 

pleasure  and  have  its  extremities, 
A  and  B,  rest  upon  MN, 

Conclusion.  The  inclosed  area 
ADBA  cannot  be  a  maximum 
unless  ^i)5  is  a  semicircle.  -M"    ^ — g 

Proof.  If  the  line  is  not  a  semicircle,  there  must  be  some 
point  £  upon  It  such  that  the  angle  ADB  shall  not  be  a  right 
angle  (§§  238,  241).    Join  DA,  DB.  ^ 

The  ^reaADBA  between  the  curve  and  the  straight  line 
IS  then  divided  into  three  parts,  which  we  may  call  AD  DB 
and  the  triangle  ADB.  '       ' 

Bend  the  curve  at  the  point  D,  leaving  the  two  branches 
AD  and  BD  unchanged,  and 
sliding  the  ends  A  and  B  along                 I^ 
the  line  MN,  so  that  the  curve 
shall  take  up  the  form  AD'B, 
in  which  AD'B  is  a  right  angle.  y^__jc .^ 

Because  the  triangles  ADB         ^  B 

and  AD'B  have  their  sides  AD  and  DB  =  AD'  and  D'B 
and  AD'B  is  a  right  angle,  ' 

Area  AD'B  >  area  ADB,  (§  488) 

while  the  other  two  areas,  AD  and  DB,  remain  unchanged. 

Therefore  the  inclosed  area  ADBA  can  be  increased  with- 
out ciianging  the  length  of  the  curve,  whence  this  area  ia  not 
a  maximum. 


m 


ff.M 


242    BOOK  VI   HMO ULAR  POLYGOm  AND  rilE  CIRCLE. 

Honco  tho  curve  cannot  inclose  a  maximum  area  unless 
AB  subtends  a  right  angle  from  every  one  of  its  points,  and 
it  is  then  a  semicircle.     Q.E.D. 


Theorem  XIV. 

494.  Of  all  areas  inclosed  by  equal  perirrieters^ 
the  circle  is  a  maximum. 

Hypothesis.     ABCD,  a  closed  line  of  given  length. 

Conclusion.  ABCD  cannot  inclose 
the  maximum  urea  unless  it  is  a  circle. 

Proof.  Take  any  two  points,  A 
and  G,  on  the  curve  so  as  to  divide  it 
into  two  equal  parts.     Join  A  G. 

Now  if  ABQ  and  ADC  are  not  a] 
both  semicircles,  suppose  the  curve- 
line  7l7?C  bent  into  a  semicircle  with- 
out changing  its  length,  the  foot  A 
remaining  unchanged  in  position,  and 
let  this  semicircle  be  AB'C\     We  shall  then  have 

Area  AB'C'A  >  area  ABGA,  (§  493) 

If  ADC  is  not  a  semicircle,  we  may  bend  it  into  the  semi- 
circle AD'C  such  that 

Area  AD'C'A  >  area  ADCA. 
Because  the  two  semicircles  are  equal  in  length,  they  are 
halves  of  equal  circles  and  the  diameters  are  equal,  so  that  the 
two  points  C  coincide. 

Adding  the  two  areas,  we  shall  have 

Area  of  circle  AB'G'D'A  >  area  ABGDA.      • 
Therefore  the  area  ABCD  wiU  not  be  a  maximum  when  it 
is  not  a  circle.     Q.E.D. 


Theorem  XV. 

495.  A  polygon  of  wMcTi  all  the  sides  are  given 
incloses  a  maximum  area  when  it  can  he  inscribed  in 
a  circle. 


TUKOUEMa  FOli  EXMIWWE. 


243 


Hypothesis,    if,  a  polygon  inscribed  in  acirclo;  N  anv 
other  polygon,  having  its  >     »      J 

sides  equal  in  length,  num- 
ber, and  arrangement  with 
those  of  the  polygon  if.     E 

Conclusion.  f 

Area  if  >  area  JV.  \ 

Proof.  1.  Upon  the 
sides  of  N  describe  arcs  of 
circles  equal  to  the  arcs  upon  the  sides  of  M.     Then 

Area  M=  area  of  circle  -  area  of  segments. 

Area  N  =  urea  of  distorted  circle  -^  area  of  segments. 

2.  Jiecausc  each  segment  around  a  side  of  A^  is  identically 
equal  to  the  segment  around  tlie  corresponding  side  of  M 
the  areas  of  the  two  sets  of  segments  are  equal.  ' 

3.  Because  the  circumferences  of  the  true  circle  around 
M  and  the  distorted  circle  around  A^are  equal. 

Area  of  circle  >  area  of  distorted  circle.       (8  494^ 

4.  Compai-ing  with  (1),  ^ 

Area  if  >  area  a:     Q.E.D. 

Corollary.  It  has  been  shown  that  the  maximum  polygon 
of  given  perimeter  and  number  of  sides  has  equal  sides. 

if  a  polygon  with  all  its  sides  equal  be  inscribed  in  a  circle. 
It  must  have  its  angles  equal  and  bo  regular  (§  447).     Hence: 

496.  A  polygon  of  wMcli  the  perimeter  and  number  of 
sides  are  given  incloses  the  maximum  area  when  it  is  regular. 

Theorems  for  Exercise. 

1.  The  inclosed  area  between  two  concentric  circles  is 
equal  to  the  area  of  a  circle  whose  diameter  is  that  chord  of 
the  outer  circle  which  is  tangent  to  the  inner  one. 

r^nlv;  J^'  T""'  ^^  \'^''^'  ''  *^  *^^*  ^*  ^^y  Circumscribed 
polygon  as  its  circumference  to  the  perimeter  of  the  polygon. 

3.  The  area  of  the  regular  inscribed  hexagon  is  a  mean 
proportiona  between  the  areas  of  the  inscribed  and  circum- 
scribed  equilateral  triangles. 

.f  ft'  "^r'  ^ffP"^?^  1'^'^"'^^^  octagon  is  equal  to  the  rectangle 
of  the  sides  of  the  inscribed  and  circumscribed  squares. 


11 


BOOK  VII. 
OF  LOCI  AND  CONIC  SECTIONS, 


CHAPTER    I. 
LINES    AND    CIRCLES    AS    LOCI. 


497.  To  fix  the  position  of  a  point  on  a  plane,  two  inde- 
pendent conditions  are  required. 

Example.  If  a  point  is  subject  to  the  condition  that  it 
must  be  two  inches  below  one  line  and  one  inch  to  the  right 
of  another  perpendicular  line,  its  position  is  completely 
fixed. 

But  if  the  only  condition  is  that  it  must  be  two  inches 
below  a  given  horizontal  line,  its  position  is  not  fixed,  but  it 
may  move  along  a  line  two  inches  below  the  given  one.  This 
last  line  is  then  called  the  locus  of  the  point. 

498.  Def,  The  locus  of  a  point  is  a  line  or  group 
of  lines  to  which  the  point  must  be  confined  when 
subject  to  some  one  condition. 

Problem  I.* 

499.  To  find  the  locus  of  a  point  which  must  he 
at  a  giten  distance  from  a  given  straight  line. 

Let   AB    be    the    given 

straight  line,  and  call  a  the  *'*"r ~ ^ 

given  distance.  :* 

Draw  MN  and  PQ  parallel  ^*""i — ; ^ 

to  ABj  at  the  distance  a  on  ;*  ^ 

either  side  of  it.  p—*.-- Q 

*  It  is  recommended  to  the  student  that,  before  beginning  to  draw 
the  iocus  in  these  problems,  he  mark  a  number  of  points  each  fulfilling 
the  required  condition,  and  continue  marking  until  he  sees  what  the 
locus  will  be. 


"> 


LINES  AND  CIRGLBS  AS  LOCI. 


246 


Every  point  on  either  of  the  lines  MN  and  PQ  is  at  the 
given  distance  a  from  the  given  line  AB  (§  129). 

It  is  evident  that  no  other  point  in  the  plane  can  be  at 
that  distance  a. 

Therefore  the  two  lines  MN  and  PQ  form  the  required 
locus  of  the  point  at  the  distance  a  from  AB. 

600.  Corollary.  If  the  condition  were  th^t  the  point 
must  be  at  the  distance  a  above  the  line  AB,  the  locus  would 
be  the  line  JfiV^  alone. 

If  the  point  must  be  at  the  distance  a  Mow  th'^  line  AB 
the  locus  would  be  P^  alone.  ' 

Problem  II. 

501.  To  find  the  locus  of  the  point  which  is  equi- 
distant from  two  given  straight  lines. 

Let  AB  and  CD  be  the  two  lines,  and  0  their  point  of 
intersection. 

Let  each  of  the  four  angles  at    ^ 
0— namely,    BOD,    DOA,  AOG, 
COB— be  bisected  by  the  respective  M 
lines  OJV,  OQ,  OM,  OP. 

Every  point  on  the   bisecting   ^ 
lines  will  be  equally  distant  from 

the  two  given  lines  (§  106),  and  every  other  point  will  be  un- 
equally distant. 

Therefore  these  bisectors  form  the  required  locus.  By 
§  85  they  form  a  pair  of  straight  lines  at  right  angles  to  each 
other. 

Therefore  the  locus  of  the  point  which  is  equidistant  from 
two  given  straight  lines  is  a  pair  of  lines  at  right  angles  to 
each  other,  bisecting  the  angles  formed  by  the  given  lines. 

Scholium.  There  are  two  ways  of  thinking  of  the  rela- 
tion of  a  point  to  its  locus  which  both  amount  to  the  same 
thing. 


1.  That  a  row  of  points,  as  numerous  and  close 


as  we 


choose,  lie  on  the  locus.     Every  one  of  these  points  will  then 
fulfill  the  given  condition. 


i.       rfi 


''i  bl' 


•B 


246       BOOK  VII.     OF  LOCI  AND  OONIG  8E0TI0m. 

2.  That  the  point  slides  along  the  locus.  The  point  wiU 
then  fulfill  the  condition  so  long  as  it  does  not  leave  the  locus. 

Examples.  1.  If  we  make  any  number  of  points  on  the 
lines  MN  and  PQ,  every  one  of  these  points  will  be  equally 
distant  from  the  lines  AB  and  CD. 

2.  If  we  slide  a  point  along  the  lines  MN  and  PQy  it  will 
always  be  as  far  from  the  line  AB  as  from  the  line  CD, 

Pjjoblem  III. 

502.  To  find  the  locus  of  the  point  subject  to  the 
condition  that  it  shall  he  equally  distant  from  two 
given  points.  tp 

Let  A  and  B  be  the  given  j 

points.  Join  them  by  a  straight  j 

lino,  and  bisect  this  line  at  OA, jO 

by  another  line  PO^  at  right  j 

angles  to  it.  \ 

Then  every  point  on  PQ  •    ! 

will  be  equally  distant  from  the 

two  points  A  and  B,  and  every  other  point  will  be  unequally 
distant  (§  104). 

Therefore  P^  is  the  locus  of  the  point  which  is  equally 
distant  from  A  and  B. 

Therefore  the  locus  of  a  point  equally  distant  from  two 
fixed  points  is  the  perpendicular  bisector  of  the  straight  line 
joining  the  points. 

Peoblem  IV 

503.  To  find  the  locus  of  the  point  which  is  at  a 
given  distance  from  a  given  point. 

Let  0  be  the  given  point,  and  a  the  given  distance. 

Around  0  as  a  centre  describe  a  circle         -^^^'"---^ 
with  the  radius  a.  /  v 

Every  point  on  this  circle  will  be  at  the  / 
distance  a  from  0,  and  every  point  either  ( 
inside  or  outside  the  circle  will  be  at  a  less  ^ 
or  greater  distance  from  0  (§  206). 

Therefore  every  point  on  the  circle  ful- 


/ 


LINES  AND  0IR0LE8  AS  LOCI. 


247 


fills  the  required  condition  of  being  at  the  distance  a  from  0 

and  no  other  point  does.  ' 

Therefore  the  locus  of  the  point  at  the  distance  a  from  the 

point  0  IS  a  circle  of  which  0  is  the  centre  and  a  the  radius. 

Problem  V. 

504.  To  find  the  locus  of  the  point  from  which  a 
given  line  suhtmds  a  right  angle. 
Let  ^  J5  be  the  given  line. 
On  ^i?  as  a  diameter  describe  the  circle  APB. 
If  P  be  any  point  on  this  circle,  the  angle  APB  will  be  a 
right  angle  (§  238).     Prom  any  point 
inside  or  outside  of  the  circle  the  angle 
will  be  greater   or    less  than  a  right 
angle  (§2*2). 

Therefore  every  point  on  the  circle  ^ 
fulfills  the  required  condition,  and  no    \  / 

other  point  does.  \^  / 

Therefore  the  locus  of    the   point         '*'* *''' 

from  which  the  line  AB  subtends  a  right  angle  is  the  circle 
described  around  ^i?  as  a  diameter. 

We  may  also  say :  If  the  point  P  slides  around  the  circle 
APBy  the  angle  APB  will  always 
be  a  right  angle. 

Corollary  1.  This  result  teaches 
us  a  curious  method  by  which  a 
circle  maybe  described.  Drive  two 
pins  A  and  B  into  the  surface  rep- 
resenting the  plane.     Take  a  common  square,  and  fasten  a 
pencil-pomt  into  its  interior  angle  P.     Then  slide  the  square 
around  on  the  two  pins,  and  the  pencil-point 
will  describe  a  circle.     The  pins  will  be  at 
the  extremities  of  a  diameter  of  the  circle. 

505.  Cor.  2.  It  may  be  shown  in  the 
same  way  ^that  the  locus  of  all  the  points 
from  which  a  given  line  subtends  a  given 
angle  different  from  a  right  angle  is  formed 
of  two  arcs  of  circles.     (Compare  §230.) 


248       BOOK  Vll.     OF  LOCIANU  CONIC  SECTIONS. 


Problem  VI. 
506.  To  find  the  locus  of  the  point  svbject  to  the 
condition  that  its  distances  from  two  given  points 
shall  have  a  given  ratio  to  each  other. 

Let  A  and  B  be  the  two  given  points,  and  let  the  given 
ratio  be  that  of  m  :  w. 

Let  P  be  any  position  of  p 

the  required  point.   Join  AB, 
PA,  and  PB.  .^ ,      ,  ^ 

Bisect  the  angle  ^P^  in-  Q      B  ^ 

ternally  by  the  line  PQ,  cutting  AB  internally  at  Q,  and 
bisect  the  adjacent  exterior  angle  by  PR,  cutting  AB  ex- 
ternally in  R. 

Then,  by  the  given  condition, 

PA  :  PB  ::m  :n. 
Therefore 

AQ  :BQ  ::m  :  n.  (§405) 

AR  :  BR  '.'.mm.  (§  406) 

Because  of  the  equality  of  these  ratios,  the  line  ABi&  cut 
hai  monically  in  the  points  Q  and  R  (§  407). 

Because  the  condition  requires  that  the  lines  PA  and  PB 
constantly  have  this  same  ratio  m  :  n,  it  follows  that  the 
bisectors  in  question  constantly  pass  through  the  same  points 
Q  and  R,  wherever  the  point  P  may  move. 

But  these  bisectors  are  at  right  angles  to  each  other  (§  82). 
Therefore  the  angle  QPR  is  a  right  angle,  and  the  locus 
of  P  is  the  same  as  the  locus  of  the  point  from  which  the  line 
QR  subtends  a  right  angle. 

Therefore  the  required  locus  is  the  circle  described  around 
QR  as  a  diameter,  the  points  Q  and  R  being  fixed  by  the 
c  nditions 

AQ  :  BQ  ::m  :  n. 
AR  :  BR  ::  m  :  n. 

Problem  VII. 
507.  To  find  the  locus  of  the  point  from  which 
two  adjacent  segments  of  the  same  straight  line  sub- 
tend equal  angles. 


LIMITS  OF  FIQUBES. 


249 


Let  AB  and  BO  be  the  two  adjacent  segments,  and  P  any 
position  of  the  point. 

By  the  condition  we  must 
have 

Angle  APB  =  angle  BFG. 
Therefore  PB  is  the  bisector 
of  the  angle  A  PC,  and  in  con- 
sequence the  lines  PA  and  PC  fulfill  the  condition 

BA:PC::AB:Ba 
Because  the  points  A,  B,  and  C  are  fixed,  the  ratio  ABiBC 
is  a  constant.  Therefore  the  ratio  PA  :  PC  is  also  a  con- 
stant, and  the  locus  is  that  of  the  point  whose  distances  from 
A  and  O  have  a  given  ratio,  AB  :  BO,  to  each  other.  This 
locus  is  a  circle  ( §  50G). 

Note.    The  locus  may  be  found  independently  of  Prob.  VI.  by 
drawing  PD  at  right  angles  to  PB,  and  then  reasoning  as  in  Prob.  VI. 


•  ♦  » 


CHAPTER    II. 

LIMITS  OF  CERTAIN   FIGURES. 


Theorem  I. 

508.  If  the  vertex  of  an  isosceles  triangle  be 
carried  away  from  the  base  indefinitely,  each  angle 
at  the  base  will  approach  a  right  angle  as  its  limit. 

Hypothesis,   ABO,  an  isosceles  triangle  in  which  CA  =  OB; 
OD,  the  perpendicular  from  the  vertex  O 
upon  the  base,  bisecting  the  latter.  f 

Oonclusion.  If  the  vertex  O  be  carried 
out  indefinitely  along  the  line  DO,  produced 
past  /,  each  of  the  angles  DBO  and  DAO 
will  approach  a  right  angle  as  their  limit. 

Proof,  Through  B  draw  a  line  BI 
parallel  to  DO,  and  therefore  perpendicular 


4.^      13  71 
tU  JJU. 


1.   If  BI  is  not  the  limit  of  BO,  let  j^ 
some  other  line,  BI',  be  that  limit. 


.«« 


260       BOOK  VII.     OF  LOGI  AND  CONIC  SECTIONS. 

2.  Because  BI'  is  not  parallel  to  DG,  it  will  meet  it  if  suf- 
ficiently produced  (§  45,  Ax.  11).   Call  the  point  of  meeting  y. 

3.  By  carrying  the  vertex  G  beyond  y,the  angle  DBG 
will  become  greater  than  DBI\ 

4.  Because  this  is  true  howeyer  small  the  angle  I'BI^  the 
angle  DBG  has  no  limit  less  than  the  right  angle  DBL 

6.  DBG  can  never  become  equal  to  a  right  angle,  because 
then  the  triangle  GDB  would  have  the  angles  D  and  B  both 
right  angles. 

6.  Therefore  the  right  angle  DBI  is  the  limit  of  the  angle 
DBGy  as  the  vertex  Cgoes  out  indefinitely  along  the  lineZ>»/. 

In  the  same  way  it  may  be  shown  that  the  limit  DAG  is 
a  right  angle.    Q.E.D. 

509.  Gorollaryl.  As  the  vertex  C  goes  out  indefinitely, 
each  of  the  sidep  BG  and  -4  C' will  approach  the  position  of  par- 
allelism to  the  perpendicular  GD  as  their  limit,  and  will  there- 
fore approach  indefinitely  near  to  parallelism  with  each  other. 

Gor,  2.  The  same  thing  being  supposed,  because  the 
angle  DGB  and  DGA  are  each  supplements  of  GAD  and 
GBDy  and  these  angles  approach  indefinitely  near  to  right 
angles,  we  conclude: 

510.  The  angle  at  the  vertex  approaches  zero  as  its  limit. 

Theorem  II. 

511.  IftJie  vertex  of  a  right-angled  triangle  he 
lengthened  out  indefinitely,  the  adjacent  side  will 
approach  the  length  of  the  hypothenuse  as  its  limit. 

Hypothesis.    ABG,  Vi  right-angled  triangle  of    . 
which  the  base  AB  is  fixed,  but  of  which  the   (] 
vertex  G  may  be  carried  out  indefinitely  along  the 
line  A  G  produced. 

Gonclusion.  However  great  the  distance  AB, 
the  vertex  G  m&j  be  carried  out  to  such  a  dii^tance 
that  the  difference  GB  -  GA  shall  be  le^s  than 
any  length  we  can  assign. 

Remabk.  We  mav  fixnrpsH  thn  pnni-»liiqi/ip  tr.  fv,?« 
form:  How  many  miles  soever  may  be  the  base  A7i,  we 
can  cany  the  vertex  C  so  far  out  that  the  exco:.  J  CB 


LIMITS  OF  FIGURES. 


251 


over  OA  shall  be  less  than  a  foot,  less  than  an  inch,  less  than  the 
hundredth  of  an  inch,  and  so  on  indefinitely. 

Proof,     From  A  draw  AD  ±  BC.     Then— 

1.  Because  CD  A  and  CAB  are  right  angles,  CA  is  greater 
than  CD  but  less  than  CB. 

2.  The  triangles  BDA  and  ^^Care  similar  (§  400). 

3.  As  the  vertex  C  moves  out  indefinitely,  the  ratio 
BA:  BC  will  approach  zero  as  its  limit.  Therefore  the  ratio 
BD  :  BA  will  also  approach  zero  as  its  limit;  that  is,  AB 
being  constant,  the  point  D  will  approach  B  as  its  limit. 

4.  Then  CA,  being  between  CD  and  CB,  will  approach 
CB  as  its  limit.     Q.E.D. 


[  - 


Theorem  III. 

512,  If  the  radius  of  a  circle  increase  indefinite- 
ly., an  arc  of  the  circle  of  gjiisen  lengtJi  will  approach 
a  straight  line  as  its  limit. 

Proof,    1.  Let  RT  be  the 
length  of  the  given  arc. 

2.  At  R  erect  the  line  RO 
perpendicular  to  RT,  and  join  x 
OT. 

3.  From  0  as  a  centre  describe  an  arc  equal  to  72  r  and 
passing  through  R. 

4.  Let  the  centre  0  move  out  indefinitely  along  the  line 
RO  produced,  the  circle  still  passing  through  R,  so  that  OR 
is  its  radius. 

5.  If  K  be  the  point  in  which  the  circle  intersects  the 
hue  OT,  we  sliall  have  OK  =  OR.  Therefore,  as  0  moves  out 
mdefinitely,  the  point  K  m\\  approach  T^as  its  limit  (§  511), 
and  the  arc  of  the  circle  will  approach  the  straight  line  RT 
as  its  limit.     Q.E.D. 

Theorem  IV. 

£^13.     Tf  P.n.n7}.    n-f  hnn    1/inne>o    nnJinAh    /JA-4P^^  J.-.   «    

slant  quantUy,  are  extended  indefinitely,  their  ratio 
will  approach  unity  as  its  limit. 


m 


if  f 
III 


¥ 


\  m\ 


252       BOOK  VII.    OF  LOCI  AND  CONIC  SECTIONS. 

Proof.  Call  A  the  lesser  line;  />,  the  constant  difference 
of  the  two  lines;  A  -\-D,  the  greater  line;  r,  the  ratio  of 
-4  +  />  to  A  ;  a,  the  quantity  by  which  r  exceeds  unity. 
Then  A  +  D-rA  -  (I -^  a)A  =  A-^  aA, 

Therefore  D  =  aA, 

or  D  :  A  ::  a  :  1, 

Now,  we  can  increase  A  so  that  the  ratio  D  :  A  shall  be 
less  than  any  quantity  we  may  assign.  Therefore  a  may  be 
made  less  than  any  assignable  quantity,  whence  r  may  differ 
from  unity  by  less  than  any  such  quantity;  that  is,  the  limit 
ofrisl.    Q.E.D. 


>  >  « 


CHAPTER     III. 

THE  €LUPSE. 


614.  Def.  An  ellipse  is  the  locus  of  the  point, 
the  sum  of  whose  distances  from  two  fixed  points  is  a 
constant.  Each  of  the  two  fixed  points  is  called  a 
focus  of  the  ellipse. 

515,  To  describe  an  ellipse.    Let  B  and  Fhe  the  foci. 

Take  a  thread  of  which  the 
length  shall  be  equal  to  the  sum  of 
the  distances  of  each  point  ">f  the 
curve  from  the  foci,  which  sum  is 
supposed  to  be  given,  and  fasten 
one  end  in  each  focus. 

Stretch  the  thread  tight  by 
pressing  a  pencil-point  against  it,  and  move  the  latter  round, 
keeping  it  pressed  against  the  thread.  The  pencil-point  will 
describe  an  ellipse. 

Proof.  Let  P  be  any  point  of  the  curve  described  by  the 
pencil-point.  The  sum  of  the  distances  of  this  point  from 
6UU  ioui  IB  x-jd  -j-  s-jr,  Dui  x'ji,  -f-  I'^f  makes  up  the  whole 
length  of  the  thread  which  remains  constant.    Therefore  the 


THE  ELLIP8EL 


263 


1b 


sum  of  the  distances  of  P  from  the  foci  is  equal  to  this 
constant,  whence  P  is  by  definition  a  point  of  the  ellipse. 

616.  Axes  of  the  ellipse.  In  drawing  the  ellipse  there 
will  be  two  points,  A  and  /?,  where  the  two  parts  of  the  thread 
will  overlap  each  other.  The  line  AB  is  called  the  major 
azifl  of  the  ellipse. 

Let  us  put  I  =  the  length  of  the  thread.    Then 

AB-{-AF=l;l  /^ 

£!B-\-FB  =  l\  ^^^ 

Adding  these  equations,  and  noting  thai  AF  =  AB 4-  EF 
and  EB-EF-\-  FB,  we  have 

%AE-\-'iEF-^%FB  =  'U, 
Dividing  this  equation  by  3, 

AE-^EF'\-FB=il, 

^"^  AB  =  l     Hence: 

61 7.  The  major  axis  of  the  ellipse  is  equal  to  the  sum  of 
the  distances  of  each  point  of  the  ellipse  from  the  foci. 

The  same  equations  (a)  also  give 

AE-\-AF=EB  +  FB, 
or  2AE-{-EF=:EF-i-.2FB; 

whence  AE=FB, 

Hence  the  foci  are  equally  distant  from  the  ends  of  the 
major  axis. 

If  0  be  the  middle  point  of  the  major  axis,  the  distance 
OA  is  called  the  semi-major  axis,  and  is  represented  by  the 
letter  a.     Then  2a  is  the  major  axis;  whence 
2a  =  I,  the  length  of  the  string. 

618.  Minor  axis.    In  drawing  the  ellipse  there  will  be 
two  points,  G  and  J9,  equidistant 
from   the   two   foci.     Join    these 
points  by  the  line  CD,  intersecting 
the  major  axis  in  0.  Af 

Because,  in  the  quadrilateral 
ECFD,  CE  =  ED  =  DF=:  FC,  this 
quadrilateral  is  a  rhombus,  and  the 
diagonals  J^and  CD  bisect  each  other  at  right  angles  (§173). 

Hence  CD  is  the  perpendicular  bisector  of  the  major  axis 
AB,     CD  is  called  the  minor  axis  of  the  ellipse. 


254        DOOK  VII.     OF  LOCI  AND  CONIO  SECTIONS, 


519.  Dtf.  The  minor  axis  of  the  ellipse  is  that 
segment  of  the  perpendicular  bisector  of  the  major 
axis  which  is  terminated  by  the  curve.      ,  v 

Also: 

The  point  0  is  called  the  centre  of  ^he  ellipse. 

The  major  and  minor  axes  aro  callea  the  principal 
axes  of  the  ellipse. 

The  distance  of  the  centra  ^  r^om  each  of  the  foci 
is  called  the  linear  eccentricity  of  the  ellipse. 

The  ratio  of  the  linear  eccentricity  to  the  semi- 
major  axis  is  called  the  eccentricity  of  the  ellipse. 
That  is, 

Eccentricity  =  OE :  OA  =  — . 

520,  It  is  common  to  use  the  notation: 
by  the  semi-minor  axis  of  the 

ellipse  =  00. 

c,  its  linear  eccentricity. 

e,  the  eccentricity  of  the  elhpse.  A 

The  relation  between  the  two 
eccentricities  is  then  expressed  by 
the  equations 

e  =  —  (because  c  =  OJS). 

c  =  ae. 

,  621.  To  find  the  length  of  the  minor  axis  of  an  ellipse. 
Because  BOO  is  a  right-angled  triangle,  and  BO  =  a, 
b'  =  a'~  c'  =a'  (1  -  e'). 
Whence,  by  extracting  the  square  root, 

b=a  Vr^^, 
which  enables  us  to  determine  the  length  of  the  minor  axis 
when  the  major  axis  and  the  eccentricity  are  known. 

Note.    In  the  older  geometry  the  length  OB,  which  we  have  called 
the  linear  eccentricity,  was  called  the  "eccentricity"  simply.    But  the 


i^-r^iCi  ii3     use     lliC 


'orvi  eeeestiieitj  tu  uumguuic  mc  ratio  oi  tiuo  lengta 


OE  to  tlie  major  axis,  according  to  the  above  definition. 


THE  ELLIPSE. 


B 


m 


255 


an  ellipse  is  any  straight 


Tangent 


522.  D^,  A  chord  of 

line    terminated    by    two 
points  of  the  ellipse. 

523.  Def,  A  diameter 
of  an  ellipse  is  any  chord 
passing  through  the  cen- 
tre. 

R9>±      A   *«_  .     i  Diameter.        Chord. 

Theoeem  v. 
.^^^'9^^^!/  point  without  the  ellipse  the  svrn 

ellipse. 

Conclusions.  I.  PE  -\-  PF  >  2a. 
11.   QE+QF<  2a. 

Proof.  I.  Let  T  be  the  point  in 
which  the  line  BP  intersects  the 
ellipse.    Join  TF.     Then— 


^P  +  PF=  FT  4-  TP  4-  PF 
TP-^PF>TF 

EP  -\-PF>  FT-^TF. 
ET+TF^U. 


(Def.  of  ellipse.) 


1. 
^erefore 

2. 

Therefore 

TT    I,    .     ^P  +  PP>2a.     Q.E.D. 

J^72     T^         '  ^^  ^^*^^  '^  ^«^*«  *^e  ellipse  in  P.     Join 
^•ff.    Then  we  prove,  as  in  (I.),  ^  *^°^^ 

EQ  +  QF<FR-\.RF, 
ER  4-  ;?;?»—  o« 

Whence  "     '  "'^'  ~ '^"• 

EQ+QF<2a.    Q.E.D. 


E(.    r  t 


H, 

•!l. 


■ffl 


f'fl 


<-! 


11 


266       BOOK  VII    OF  LOCI  AND  CONIO  JSEVTI0N8, 


Theorem  VI. 

526.  If  through  any  point  of  an  ellipse  we  d/raw 
a  line  making  equal  angles  with  the  lines  froui  that 
point  to  thefoci^  that  line  will  he  a  tangent^  and  the 
only  tangent^  to  the  ellipse  at  that  point. 

Hypothesis,   E,  Fj  foci  of  an  ellipse;  P,  any  point  of  the 
ellipse;    TPF,    a    straight   line 
through  P  such  that 
Angle  TPE  =  angle  VPF, 

Conclusion.  TP  V  will  be  a 
tangent  to  the  ellipse,  and  every 
other  line  passing  through  P  will 
intersect  the  ellipse. 

Proof,   1.  Because  P  is  a  point  of  the  ellipse, 

BP-{-PF=2a, 

2.  Because  angle  TPB  =  angle  VPF,  the  sum  of  the 
distances  from  P  to  the  foci,  that  is,  FP  -{-  PF,  or  2a,  is 
less  than  the  sum  of  the  distances  from  any  other  point  of 
TFto  the  foci  (§490). 

3.  Because  the  sum  of  the  distances  of  every  other  point 
of  TFfrom  the  foci  is  greater  than  2a,  every  such  other  point 
is  without  the  ellipse  (§  625). 

4.  Therefore  the  line  TV  touches  the  ellipse  at  P  without 
intersecting  it,  and  is  therefore  a  tangent.     Q.E.D. 

6.  If  any  other  line  than  TP  V  passes  through  P,  it  can- 
not make  equal  angles  at  P  with  the  lines  PF  and  PF. 
Therefore  there  will  be  some  point  of  the  line  for  which  the 
sum  of  the  distances  from  F  and  F  will  be  less  than  FP  + 
PF{^  490);  that  is,  the  line  will  pass  inside  of  the  ellipse  and 
cannot  be  a  tangent.     Q.E.D. 

537.  Bef.  Two  points  so  situated  that  an  in- 
definite line  is  the  perpendicular  bisector  of  the  line 
joining  them  are  said  to  be  opposite  points  with 
respect  to  that  indefinite  line. 

of  the  line  joining  two  opposite  points  with  respect  to  it. 


TIIK  KLLWHE. 


257 


raw 
that 
I  the 

:  the 


the 
Uj  is 
it  of 


)oint 
>oiut 

bout 

can- 
Pi^. 

the 

P  + 
and 


in- 
line 
vlth 


(§  104) 

(§  102) 

(§67) 


Theorem   VII. 
528.  Tlie  line  from  any  focus  to  the  opposite 
point  of  the  other  focus,  with  respect  to  a  tangent 
passes  through  the  point  of  tangency,  ' 

Hypothesis.     VT,  a  tangent  to  an  ellipse  having  E  and  F 
as  foci ;  F",  the  opposite  point 
of  F  with  respect  to  the  tangent;  ^- 

P,  the  point  in  which  the  lino 
iS'iF*  intersects  VT. 

Conclusion.  P  is  the  point  of 
tangency. 

Proof,    1.  Because  Fr  is  the 
perpendicular  bisector  of  FF'y 

PF  =  PF. 
Angle  TPF  =  angle  TPF' 

«    t>  .  =  angle  ^Pr.  ,^  „., 

2.  Because  the  angles  FPT  and  FPVaro  equal,  P  is  the 
point  at  which  the  tangent  touches  the  ellipse  (§526).  Q.E.D. 

Corollary.     Because  P  is  a  point  of  the  ellipse,  we  have 

^P  +  PP=2a, 
and  because  PP'  =  PF,  we  have  also 

FF'  =  2a. 
Therefore  the  opposite  point  of  any  one  focus  is  at  the  dis- 
tance 2a  from  the  other  focus,  and  we  conclude: 

529.  The  locus  of  the  opposite  point  of  one  focus,  with 
respect  to  a  moving  tangent,  is  a  circle  around  the  other  focus 
with  the  radius  2a. 

In  other  words,  if  a  tangent  roll  round  on  an  eHipse,  the 
opposite  point  of  either  focus  will  describe  a  circle  round  the 
other  focus  as  a  centre  with  the  radius  2a. 

530.  This  theorem  and  corollary  afford  an  elegant 
method  of  drawing  any  number  of  tangents  to  an  ellipse  with- 
out drawing  the  ellipse  itself.  We  need  only  to  know  the 
positions  of  the  foci  and  the  length  of  the  major  axis. 

Construction.     Let  F  and  F  be  the  given  foci. 

Around  either  focus,  as  F,  with  a  radius  equal  to  the 


11 

iffl 

1 

^IH 

■*l£^H 

UHH 

T' 

1  ^^^1 

■  : 

il 

' 

WM 

1  i 

SBKBU 

1 

^^^H 

?  * 

.    ! 

j||^B 

WM 

/d^J^ 


268       BOOK  VII.    OF  LOOI  AND  OONIO  SEOTIONS. 


major  axis  we  describe  a  circle.    This  circle  will  then  be  the 
locus  of  all  the  opposite  points  of  F 
with  respect  to  the  tangents. 

We  draw  any  line  from  F  to  the 
circle,  and  bisect  this  line  at  right 
angles  by  another  line. 

The  bisecting  line  will  be  a  tan- 
gent to  the  ellipse. 

By  drawing  a  number  of  such 
lines  any  number  of  tangents  may  be 
drawn. 

Theorems  for  Exercise. 
I.  Each  principal  axis  of  an  ellipse  is  an  axis  of  symmetry. 
II.  The  ellipse  is  symmetrical  with  respect  to  its  centre  as 
a  centre  of  symmetry. 
III.  Every  diameter  is  bisected  at  the  centre. 
rV.  The  tangents  at  the  two  ends  of  a  diameter  are  parallel. 


•  ♦ 


CHAPTER   IV. 
THE    HYPERBOLA. 


631.  Def,  An  hyperbola  is  the  locus  of  the  point 
the  difference  of  whose  distances  from  two  fbced 
points  is  a  constant. 

Each  of  the  two  fixed  points  is  called  a  focus  of  the 
hyperbola. 

63S.  Any  number  of 
points  of  an  hyperbola  may  be 
found  by  the  intersection  of 
two  circular  arcs,  thus: 

Let  2a  be  the  constant  dif- 
ference between  the  distances 
of  a  point  of  the  curye  from 
the  foci.     From  either  focus  as  ^ 
a  centre,  with  an  arbitrary  radius  r  describe  an  arc  of  a  circle. 


THE  HYPERBOLA. 


259 


the 


ry. 
as 


iel. 


nt 
ed 

he 


le. 


From  the  other  focus,  with  the  radius  r  +  2a,  describe  an 
arc  intersecting  the  other  arc. 

The  point  of  intersection  will  be  at  the  distance  r  from 
one  focus  and  r  +  2a  from  the  other;  the  difference  of  those 
distances  is  2a,  whence  the  point  lies  on  the  hyperbola. 

633.  Corollary.  Since  there  is  no  limit  to  the  radius  r, 
the  hyperbola  extends  out  to  infinity. 

534.  Major  axis  of  the  hyperbola.  If  the  constant  2a 
were  greater  than  the  distance  between  the  foci  F  and  F', 
there  would  be  no  point  the  difference  of  whose  distances 
from  i^and  F'  could  be  as  great  as  2a,  and  so  there  would  be 
no  hyperbola.    Therefore  2a  must  be  less  than  FF'. 

Again,  if  we  pass  along  the  line  FF'  from  F  to  F\  the 
difference  of  the  distances  will  be  FF'  when  we  start,  it  Will 
diminish  to  zero  at  the  middle  point  of  the  line,  and  will  then 
increase  to  FF'  at  the  end  F'.  Hence  there  must  be  two 
points  on  the  line  for  which  this  difference  is  2rt;  that  is,  two 
points  of  the  hyperbola.  Let  A  and  B  be  these  points.  We 
must  then  have,  by  the  conditions  of  the  locus, 

BF-  BF'  =  2a;  that  is,  FA  +  AB  -  BF'  =  2a. 

AF'  -AF=  2a;  that  is,  ^  FA  +  AB -\^  BF'  =  2a. 
The  sum  of  these  equations  divided  by  2  gives 

m,   .  AB  =  2a. 

Their  difference  gives 

FA  =  BF'. 
From  these  two  equations  we  readily  see  that  the  curve 
cuts  the  line  FF'  at  the  distance  a  on  each  side  of  the  middle 
point  0  of  that  hue. 

535.  Def.  The  distance  between  the  points  at 
which  the  hyperbola  cuts  the  line  joining  its  foci  is 
called  the  major  axis  of  the  hyperbola. 

From  what  has  been  said  we  see  that  the  major  axis  is 
equal  to  the  common  difference  of  the  di&lances  from  each 
point  of  the  curve  to  the  foci. 

Since  a  point  of  the  hyperbola  may  be  either  nearer  to  F 
than  to  F'  by  2a,  or  nearer  to  F'  than  to  F,  the  hyperbola 
consists  of  two  branches. 

Also,  if  we  draw  the  perpendicular  bisector  of  FF'  through 


S    tl 


Um       BOOK  VII.    OF  LOCI  AND  GONIG  8EGTI0N8. 


0,  every  point  of  this  bisector  being  equally  distant  from  F 
and  F'y  no  point  of  it  can  be  a  point  of  the  hyperbola:  Hcljo 

536.  The  two  branches  of  the  hyperlola  are  completely 
separated. 

Remark  1.  Most  of  the  properties  of  the  ellipse  and 
hyperbola  correspond  to  each  other  in  that  where  one  has 
sums  of  lines,  the  other  has  differences;  where  an  angle  is 
formed  in  one,  the  adjacent  angle  will  be  formed  in  the  other, 
etc.  The  student  should  compare  the  corresponding  theorems. 
Remark  2.  Since  each  branch  of  the  hyperbola  extends 
out  tc  infinity,  it  may  be  considered  as  dividing  the  plane  into 
three  distinct  parts,  one  within  each  branch  and  one  between 
the  branches.  The  two  first  portions  may  be  considered  as 
belonging  to  one  class,  and  as  lying  within  the  hyperbola— 
i.e,y  within  one  of  its  branches— and  the  last  as  lying  without 
the  hyperbola. 

Theorem  VIII. 

olH.  From  every  point  without  the  hyperbola  the 
difference  of  the  distances  from  the  foci  is  less  than 
the  major  axis. 

From  every  point  within  the  hyperbola  that  differ- 
ence is  g,  eater  than  the  major  axis. 

Hypothesis.  E,  F,  foci  of  an 
hyperbola;  P,  any  point  without 
the  hyperbola;  Q,  any  point 
within  the  hyperbola. 

Conclusions. 

I.  PF-PF<2a. 
II.   QE-QF>  2a. 

Proof.  I.  Let  N  be  the  point 
in  which  the  line  PF  intersects 
tlie  curve,  and  call  ^  the  amount  by  which  PF  exceeds  PF 
Then  A  =  PE-PF=PE-  PN-  NF, 

Because  N  is  on  the  curve, 

9,/»  — •    JP.W  —   ATJST 

Because  PE  is  a  straight  line, 

EN  4-  PN  >  PE',  whence  PE  -  PN  <  EN. 


' 


THE  HYPERBOLA. 


961 


' 


1^ 


Therefore 

PE-^PN-NF<  MN -  NF,  or  J  <  "Ja.     Q.E.D. 

II.  Let  M  be  the  point  in  which  QE  first  intersects  the 
curve,  and  call  A  the  excess  of  QE  over  QF     Then 

A  =  QE~QF, 
Because  if  is  on  the  curve, 

U=zME-MF 

=:QE-  {MQ  +  MF), 
Because  ^i^is  a  straight  line, 

MQ-\-MF>QF. 
Therefore  to  form  A  we  take  from  QE  a  less  line  than  we  do 
to  form  2a,  whence 

A  >  2a.     Q.E.D. 

538.  Corollary.  Since  every  point  on  the  plane  must  be 
either  within  the  hyperbola,  without  it,  or  upon  it,  we  con- 
clude that,  conversely: 

Every  point  the  difference  of  wliose  distances  from  the  foci 
is  less  than  %a  lies  without  the  hyperbola. 

Every  point  the  difference  of  whose  distances  from  the  foci 
is  greater  than  2a  lies  within  the  hyperbola, 

539.  Problem.  Having  a 
straight  o^ne  passing  between  the 
foci,  it  is  required  to  find  that 
point  upon  it  at  which  the  differ- 
ence of  the  distances  from  the  foci 
shall  be  the  greatest. 

Solution.  Let  E  and  F  be  the 
foci;  MN,  the  line;  F,  the  focus 
nearest  the  line. 

Let  F'  be  the  opposite  point 
of  F  relatively  to  the  line,  and 
let  EF'  produced  intersect  the  line  MN  on  P. 

Let  Q  be  any  point  at  pleasure  on  tne  line, 
difference  of  the  distances.     Then- 


Call  A  the 


whence 


QF=  QF\ 
A=  QE-  QF=QE-  QF' 


i  I 


1    lit 

1       1  ! 


262       BOOK  VII    OF  LOOI  AND  CONIC  SECTIONS. 

2.  So  long  as  g  is  at  any  point  of  the  line  except  P,  QEF 
will  form  a  triangle,  and  we  shall  have 

A^qE^qF'  <EF\ 
But  when  q  coincides  with  P,  we  have 

A-PE-  PF'  =  EF\ 
Because  A  is  equal  to  EF'  at  P,  ard  less  than  EF*  at  every 
other  point  of  the  Hne,  we  conclude  that  P  is  the  point  of 
maximum  difference  of  distance. 

Because  Pand  F'  are  opposite  points,  we  have      * 

Angle  EPq  =  angle  FPq. 
Therefore  the  point  of  maximum  difference  of  distances  is 
that  from  which  lines  to  the  foci  make  equal  angles  with  the 
line. 

At  this  point  the  line  will  be  the  bisector  of  the  angle  EPF. 


Theorem  IX. 

540.  Iffrmn  any  point  of  an  hyperhoU  lines  he 
drawn  to  the  foci,  the  bisector  of  the  angle  between 
those  lines  will  be  a  tangent  to  the  hyperbola. 

Hypothesis.     E,  F,  foci  of  an  hyperbola;  P,  any  point  of 
the  curve;    PT,   the  bisector  of 
the  angle  EPF. 

Conclusion.  PT"  is  a  tangent 
to  the  hyperbola  at  the  point  P. 

Proof.  1.  Pecause  Pr  bisects 
the  angle  EPT,  the  difference  of 
distances  from  the  foci  is  less  at 
every  other  point  of  PTthan  it  is 
at  P  (§  539). 

2.  Because  P  is  on  the  curve,  the  difference  is  there  equal 
to  2a.    Hence  it  is  less  than  2«  at  every  other  point  of  the  line. 

3.  Therefore  every  other  point  of  PT  except  P  is  without 
the  eurve,  so  that  PT  touches  the  curve  a,t  P  (§  538).  Q.E.D. 
^  Scholium.  Comparing  this  theorem  with  §  526,  we  see 
t-u»t  Willie  m  the  hyperbola  the  tangent  bisects  ^he  interior 


•  \ 


THE  HYPERBOLA. 


263 


angle  of  the  triangle  £JFF,  in  the  ellipse  it  bisects  the  ex- 
terior adjacent  angle.  Therefore  if  through  the  poi>^t  P 
both  an  ellipse  and  an  hyperbola  be  passed,  having  L  and 
/"as  the  foci,  the  tangents  to  the  two  curves  at  P  will  be  per- 
pendicular to  each  other  (§  82)." 

541.  Def.  The  ellipse  and  hyperbola  are  called 
the  conio  sections,  or  simply  oonics. 

543.  Def.  Confooal  conies  are  those  which  have 
the  same  foci. 

543.  Def.  A  family  of  confocal  conies  means  an 
indefinite  number  of  conies  having  the  same  foci. 

544.  Scholium.  Two  curves  which  intersect  are  said  to 
cut  each  other  at  an  angle  equal  to  the  angle  between  their 
tangents  at  the  point  of  intersection. 

The  reason  of  this  appellation  is  that  a  curve  at  any  point 
is  considered  to  have  the  same  direction  as  its  tangent  at  that 
point. 


;  If 


I   ' 


Theoeem  X. 

545.  In  a  family  of  confocal  ellipses  and  hyper- 
holas,  all  the  ellipses  cut  all  the  hyperbolas  at  right 
angles. 

Proof.     Let  P  be  any  point  of  intersection  of  an  ellipse 
with  a  confocal  hyperbola,  and 
PK  PF  the  lines  from  P  to 
tiic  foci. 

Because  P  is  a  point  of  the 
ellipise,  the  tangent  to  the  ellipse 
at  P  bisects  the  exterior  angle 
at  P. 

Because  P  is  a  point  of  the 
ie   laiijgpnii    co   ine 
hyprrbola  at  P  bisects  the  interior  angle  EPF, 


1 1 


264       ^OOK  Vtl.    Otr  LOCI  AND  CONIG  8ECTI0N8. 


Therefore  these  tangents  are  perpendicular  to  each  other 
(§83),  and  the  curves  inter- 
sect at  right  angles  (§  644). 

546.  Asymptotes  of  the 
hyperlola.  Because  the  tan- 
gent to  the  hyperbola  at  the 
point  P  bisects  the  angle 
FPE,  it  divides  the  line 
EF  between  the  foci  at  the 
point  Q  into  two  such  seg- 
ments that 

Now  suppose  the  point  P  to  move  out  upon  the  hyperbola 
to  infinity,  j'he  ratio  EP  :  FP  will  then  approach  unity  as 
its  limit,  because  the  difference  between  its  terms  is  the  finite 
quantity  2«,  while  each  term  may  increase  to  infinity  (§  613). 

Therefore  the  point  of  intersection  Q  approaches  the 
centre  0  as  its  limit;  and,  using  the  convenient  language  of 
infinity,  we  may  say: 

547.  A  tangent  to  the  hyperbola  at  infinity  passes 
through  the  centre  of  the  hyperbola. 

548.  Be/.  The  tangents  at  infinity  are  called 
asymptotes  of  the  hyperbola. 

549.  2h  construct  the  asymptotes.  From  E  and  F  let 
the  lines  ER  and  FS,  parallel  to  the  asymptotes,  be  drawn. 

As  the  point  P  moves  along  the  hyperbola  to  infinity, 
EP  and  FP  will  approach  EP  and  FS  as  their  limits  (§  508). 

On  the  lines  EE  and  EP  take  segments  EM  and  EM^ 
each  equal  to  2a;  then,  since  PE-PF-  2a,  we  have  PM'= 
PF,  so  that  the  triangle  M'PF'is,  isosceles  and  angle  M'  = 
angle  F.  Therefore,  as  P  recedes  to  infinity,  the  point  M' 
approaches  iJf  as  its  limit,  the  angles  M'  and  i^both  approach 
right  angles  as  their  limit  (§  508).  The  triangle  EMF  is 
therefore  right-angled  at  M.  Hence  the  direction  of  the 
asymptote  is  found  thus: 

On  the  line  EF  as  a  base  erect  a  right-angled  triangle 
EMF,  of  which  the  side  ^Jf  shall  be  equal  to  2a. 


THE  HTPEBBOLA. 


265 


lii 


The  asymptote  will  be  the  line  through  the  centre  of  the 
hyperbola  parallel  to  the  side  EM. 

To  consfruct  the  required  triangle  we  notice  that,  since 
EMFis  a  right  angle,  the  vertex  M  lies  on  the  circle  of  which 
-fi^^is  the  diameter.    Hence  the  construe-,  ,  », 

tion:  ^  ^^ 

On  EF  as  a  diameter  describe  a  circle.      . 

From  either  ^  or  ^as  a  centre,  with  a  g^ 
radius  2a  describe  arcs  of  a  circle  cutting 
the  first  circle  in  the  points  if  and  M\ 

3om  EMfm^  EM\  ^       "IT^M'       \ 

Through  the  middle  point  0  of  the  circle  draw  lines  paral- 
lel to  ^Jf  and  J^if'.  ^ 

These  lines  will  be  asymptotes  of  the  hyperbola  of  which  E 
and  i^are  the  foci,  and  2a  the  major  axis. 

550.  It  is  easy  to  prove  that  if  we  draw  the  arcs  of 
circles  around  the  centre  F, 
the  chords  then  found  will  be 
parallel  to  EM  and  EM',  and 
will  therefore  lead  to  the  same 
pair  of  asymptotes. 

We  therefore  conclude  that 
the  asymptotes  of  the  i%y;pp,rhola 
consist  of  a  pair  of  straight 
lines,  intersecting  each  other  in 
the  centre  of  the  figure* 


266       BOOK  VIL    OF  LOOI  AND  OONIO  8EUTI0Jf& 


CHAPTER    V. 
THE    PARABOLA. 


661.  Definition.    A  parabola  is  the  locus  of  a 
point  equidistant  from  a  fixed 
point  and  a  straight  line.  r 

553.  Def.  The  fixed  point  is 
called  the  focus  of  the  parabola. 

553.  Def.  The  straight  line  ^ 
is  called  the  directrix  of  the 
parabola. 

554.  Def.    A  straight   line 

through   the  focus,  and  perpen-     ,^,  distances  PE  and  PF  ar. 

dicular  to  the  directrix,  is  called  ®^"*^  ^^  whatever  point  of  the 

thp  a■«^i9  of  fho -noTKiKnlQ  curve  P  may  be  placed.    AF  iu 

tiie  axis  01  me  paraOOla.  the  axis  of  the  parabola. 

Remark.  Since  there  is  no  limit  to  the  possible  distance 
of  a  point  from  both  the  focus  and  directrix,  every  parabola 
extends  out  to  infinity. 

Theoeem  XI. 

555.  I.  Mery  point  without  the  parabola  is 
nearer  to  the  directrix  than  to  the 
focus. 

11.  Every  point  within  the  para- 
hola  is  nearer  to  the  focus  than  to 
the  directrix. 

Proof  I.  Let  P  be  a  point  without  the 
parabola.  The  line  from  this  point  to  the 
focus  must  then  intersect  the  curve.  Let  S' 
Q  be  the  point  of  intersection.  Drop  the 
perpendiculai's  QS  upon  the  directrix  and 
^2^  upon  Pi?.   Then— 


R 

S 


THE  PARABOLA. 


Because  Q  is  on  the  parabola, 
Adding  PQ  to  these  equal  lines 


267 


also. 


Because  QTP 


==m^PQ; 

PR=:TRJ^TP, 


is  right-angled  at  T, 
PQ  >  TP, 
Therefore  pp  >  pj^    Q  E  D 

Proof  II.    Let  P' be  a  point  within'the  parabola.    From 
P  drop  a  perpendicular  P'S'  upon  the  directrix.    Let  oZ 
the  point  in  which  it  intersects  the  parabola     Jofn  FO' 
Then  we  prove,  as  in  the  case  of  the  ellipse  and  hypeZl  J^  ' 

8'Q'  =  Q'F, 

S'P'=:Q'P^Q'P\ 

Q'P+Q'P^>P'Fr 
whence  s'P'  ->  p'p     Q  E  D 

Corollary    Since  every  point  in  the  piane  must  be  either 
c^4tr       '  '''  "  "^*'^^*  ^'  ^^  concludVtl'a:: 

Theorem  XII. 

focus  and  the  perpendicular    \ 

upon  the  directrix  is  a  tangent  p 

^0  the  parabola,  ^ 

hypothesis.     P,  any  point  of  a^ 

anT^K^l^^^.^^^^^^^^  focus 
and  HW  the  directrix;    PP    thA 

perpendicular   upon   th^  dirttri. 


^P;i=t.S!""2l\/'-'^"^-Sle 


the 


Oonclusi 
parabola 


i-s-v    -I.- J. 


'on.     PViaa 
at  P. 


tangent  to 


j; 


! 


f\ 


m 


268       £00K  VIL    OF  LOCI  AND  OONIO  SEOTIONS. 

Proof,    Let  V  be  any  point  of  the  line  MP  V.    Join  VF, 
VR,  and  from  Fdrop  the  perpendicular  VW  on  the  direc- 
trix.    Then — 

1.  In  the  triangles  RP  V  and  FP  F, 

Angle  RPV=  angle  FP  V  (hyp.). 

PR  =  PF  {P  being  on  the  curve). 
P  V  common. 

Therefore  the  triangles  are  identically  equal,  and 

VR  =  VF. 

2.  Because  FW  is  a  perpendicular  upon  the  directrix, 

VW  <  VR; 
whence  VW  <  VF, 

and  the  point  F  is  therefore  without  the  parabola  (§  556,  II.). 

3.  Because  F  may  be  any  point  of  the  line  MP  V  except 
P,  every  point  of  this  line  except  P  is  without  the  parabola, 
and  the  line  touches  the  parabola  at  P  without  intersecting 
it.    Q.E.D. 

558.  Scholium.    The  property  of  the  ellipse  and  parab- 
ola expressed  in  this  theorem  and   in 
Theorem  VI.,  relating  to  the  ellipse,  ^^^ 

leads  to  the  use  of  these  curves  in  reflec- 
tors designed  to  bring  rays  of  light  to 

a  focus.     Since  the  curve  and  the  tan-  t':^''" — 

gent  have  the  same  direction   at  the  \'?\f?t 

point  of  tangency,  rays  of  light  are  re-    Vf^X""''" — 

fleeted  by  the  curve  as  they  would  be  by       \:V\ 

the  tangent  at  the  point  of  reflection.  ^\„.^\ 

Because  the  angles  of  incidence  and  ^v?v 

reflection  are  equal,  it  follows  that  if 
parallel  rays  of  light,  perpendicular  to  the  directrix  and 
therefore  parallel  to  the  axis,  fall  upon  a  parabolic  reflector, 
they  will  all  be  reflected  toward  the  focus. 

Conversely,  if  a  light  be  placed  in  the  focus  of  a  parabolic 
reflector,  all  the  rays  from  the  focus  _ 

will  be  parallel  after  reflection.  yf^^^^Z-^''     '^ 

In  the  case  of  the  ellipse,  thecorre-  /  /^'-/J'-— '"^""--/.M 
spending  property  leads  to  all  the  rays  ri%rvy-"-~^J»V*rJ55'/« 
which  emanate  from  one  focus  being  ^«  "^ 
reflected  to  the  other  focus. 


I 

i 

a 

f 

0 


a 

Si 

s: 
u 

t] 


THE  PARABOLA. 


269 


oin  VF, 

le  direc- 


ve). 


brix, 


»56,  II.). 
V  except 
>arabola, 
irsecting 

1  parab- 


rix  and 
eflector, 

•arabolic 


"SI 


\ 


Relations   of  the   Ellipse,  Parabola,   and 

Hyperbola. 

Theorem  XIII. 
559.  The  parabola  may  be  regarded  as  an  ellipse 
of  which  the  major  axis  is  infinite,    ' 

Proof,    Let  E  and  F  be  the  foci  of  an  ellipse,  and  A  one 
eud  of  its  major  axis. 

On  the  line  FA  produced  take  «, 
AM=EA.  -PJf  will  then  be  equal 
to  the  major  axis  (§§  616,  617). 

From  the  farther  focus  -Pas  a 
centre,  with  a  radius  FM  describe 
an  arc  of  a  circle. 

Let  P  be  any  point  of  the 
ellipse.  Join  EP,  FP,  and  pro- 
duce FP  until  it  meets  the  circle  in  R.     Then- 

1.  Because  FP  -}-  EP  =  major  axis  =  FJi,  we  have 

PE  =  PR. 
Therefore  each  point  of  the  ellipse  is  equally  distant  from  the 
focus  E  and  from  the  arc  MR, 

2.  Now  let  the  focus  F  recede  to  infinity  along  thA  line 
^^  produced. 

3.  If  MT  be  the  perpendicular  to  MA  at  M,  the  arc  MR 
will  approach  MTaa  its  limit  (§  613);  and  PR  will  approach 
parallelism  to  MF,  and  therefore  perpendicularity  to  MT  as 
its  limit  (§  509).  HencQ  each  point  P  of  the "  ellipse  will 
approach  a  position  in  which  it  is  equally  distant  from  the 
focus  E  and  the  line  MT  That  is,  it  will  approach  a  parab- 
ola having  E  as  its  focus  and  MT  as  its  directrix.    Eence: 

560.  If  one  focus  of  an  ellipse  recedes  to  infinity,  the 
ellipse  will  become  a  parabola  about  the  other  focus,     Q.E.D. 

561.  Passage  from  the  ellipse  to  the  hyperbola.    Starting 
as  m  the  last  section,  let  us  place  the 
second  focus,   F,  on  the  opposite   F"-- 
side  of  the  point  Jf  from  E;  and  let 
us,  as  before,  draw  an  arc  around 
the  centre  i^with  a  radius  FM, 

Let  FR  be  any  radius  of  this  arc; 


M    A 


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WEBSTER,  N.Y.  14580 

(716)  872-4Sd3 


270       BOOK  YII.    OF  LOOI  AND  CONIO  SECTIONS. 


and  produce  FR  to  a  point  P,  such  that  PR  =  PE,  We 
shall  then  haye  PF  —  PE  =  FR. 

Therefore  P  will  lie  upon  an  hyperbola  of  which  E  and  F 
are  the  foci,  and  FM=  FR,  the  major  axis  (§  635). 

Then,  supposing  F  to  recede  to  infinity  in  the  direction 
MF,  we  show,  as  before,  that  P  will  approach  a  parabola  of 
which  E  is  the  focus,  and  a  perpendicular  to  FM  through  M 
the  directrix. 

Scholium.  The  ellipse,  parabola,  and  hyperbola  therefore  all 
belong  to  one  class  of  curves.  It  is  shown  in  solid  geometry 
that  they  may  all  be  formed  by  the  intersection  of  a  cone  with 
a  plane,  from  which  property  is  derived  the  term  conic  section. 

Tangents  as  Limits  of  Secants. 

562.  Becafise  one  straight  line,  and  no  more,  may  be 
drawn  between  two  points,  two  points  determine  a  straight 
line  passing  through  them. 

If  the  two  points  lie  on  a  curve,  the  straight  line  passing 
through  them  is  a  secant  of  the  curve. 

663.  Def.  The  tangent  to  a  cnrve  is  the  line 
which  the  secant  approaches,  as  its  limit,  when  the 
two  points  which  determine  the  secant  come  indefi- 
nitely near  together. 

This  is  a  more  general  definition  of  a  tangent  than  that  heretofore 
given  for  the  circle,  and  applied  to  the  ellipse.  By  means  of  it  the 
fundamental  properties  of  tangents  to  the  circle  and  conic  sections  may 
be  established,  as  follows: 

564.  The  Circle.    Let  0  be  the 
centre  of  a  circle,  and  OA,  OB,  two  b| 
of  its  radii. 

Through  A  and  B  draw  a  secant.  a1 
Then  in  the  isosceles  triangle  OAB 
we  have 

Angle  A  -f  angle  B  -\-  angle  0  =  ISO"*, 
or  2  angle  A  -\-  angle  0  =  180°; 

whence  Angle  A  =  angle  P  =:   90°  ~  ^  angle  0. 

Now  let  the  T>oint  B  a'o'nrQaGh  indefiniti^l'*''  ji^**"**  ^/^  -4- 


-rx' 


--v 


««V'lVfl.        VT^     .£XB 


THE  PARABOLA. 


271 


P  Q 


The  angle  0  will  then  approach  zero  as  its  limit. 

Therefore  the  angle  A  will  approach  90°  as  its  limit,  and, 
the  tangent  being  the  limit  toward  which  the  line  ^5  ap'- 
proaches,  must  be  perpendiculai-  to  OA. 

We  thus  arrive  at  the  property  of  the  tangent  demonstrated 
in  §  226. 

565.  llie  Ellipse.   Let  E  and  F  be  the  foci  of  an  ellipse 
and  P  and  ^  two  points  .upon  it.  ' 

Through  P  and  Q  pass  a  secant. 
Join  PE,  PF,  QE,  QF. 

Now  let  the  point  Q  approach 
indefinitely  near  to  P,  and,  as  the  E 
approach    becomes    nearer,   let   us 
look  at  P  and  Q  through  a  microscope  of  which  the  magnify- 
ing power  may  increase  indefinitely. 

Then  at  the  limit,  because  the  angles  at  E  and  F  become 
zero,  EP  and  EQ,  as  also  FP 
and  FQ,  will  seem  parallel  in 
the  microscope. 

From  P  drop  the  perpen- 
dicular PR  upon  EQ,  and  from 
Q  the  perpendicular  QS  upon 
FP,    Then— 

1.  EQ-EP=zRQ. 
FP-FQ  =  8P, 

From  the  fundamental  condition  of  the  ellipse, 

EQ-\-FQ^EP-\-FP, 
we  haye  EQ-EP=zFP-^  FQ, 

whence  RQ  =  SP. 

2.  Because  the  right-angled  triangles  PRQ  and  P8Q  have 

Hypothenuse  PQ  common. 

Side  QR  =  side  P8, 
they  are  identically  equal,  and 

Angle  PQR  =  angle  QPS, 
But  PQR  is  the  angle  which  the  tangent  makes  with  the  line 
EQ  to  the  focus  at  E,  and  QP8\s  the  angle  with  PP.     Be- 
cause at  the  limit  EP  \\  EQ  and  PP||  QF,  the  tangent  PQ 
iuakes  equal  angles  with  the  lines  to  the  foci. 


(§511) 


!=»,!=" 


272       BOOK  VII    OF  LOCI  AND  CONIO  SECTIONS. 

566.  The  Hyperbola.  The  same  reasoning  will  apply  to 
the  hyperbola,  except  that  the 
foci  E  and  F  will  be  on  opposite 
sides  of  the  tangent,  and  in  con- 
sequence the  tangent  will  bisect 
the  angles  to  the  foci. 

In  the  case  of  the  parabola  the 
same  reasoning  will  always  apply, 
the  lines  UP  and  BQ  being  re- 
placed by  perpendiculars  to  the 
directrix. 


♦  • 


CHAPTER   VI. 

represe'ntation  of  varying  magnitudes  by 

CURVES. 


567.  The  changes  of  a  varying  magnitude  may  be  repre- 
sented to  the  eye  by  a  curve  on  a  system  which  we  shall 
illustrate  by  showing  the  changes  of  the  National  Debt  of  the 
United  States  between  1860  and  1880. 

In  the  following  figure  the  horizontal  line  WJT  is  divided 
up  to  represent  the  different  years.    On  tl    Tiiddle  of  each 


5/i 

/ 


/ 


wi=^ 


r--'r 


00 

s 


year  we  erect  a  perpendicular  proportional  to  the  magnitude 
of  the  debt  at  that  time,  as  given  in  millions  of  dollars  on 
each  perpendicular.     Then  by  noting  the  length  of  these  per- 


CURVES. 


273 


>ply  to 


/ 


BY 


repre- 
I  shall 
of  the 

ivided 
I  each 


nitade 
ars  on 
le  per- 


pendiculars we  have  the  different  magnitudes  of  the  debt 
presented  to  the  eye. 

These  perpendiculars  are  called  ordinals.  f 

The  ordinates  only  show  what  the  debt  was  on  July  1st  of 
each  year,  and  we  may  wish  to  know  what  it  was  at  other 
times,  supposing  it  to  have  varied  in  a  regular  way.  This  we 
do  by  drawing  a  curve  through  the  tops  of  all  the  ordinates. 
Then  the  height  of  the  curve  above  the  base  WJT  will  show  the 
length  of  the  ordinate,  and  therefore  the  amount  of  the  debt. 
Having  drawn  the  curve,  we  may  erase  the  ordinates 
entirely,  and  get  the  amount  of  the  debt  at  any  time  by 
measuring  the  height  of  the  curve  above  the  base  line  at  the 
point  corresponding  to  that  time. 

668.  Def.  A  curve  showing  the  magnitude  of  a 
varying  quantity  at  any  time  is  caUed  a  graphic 
representation  of  that  quantity. 

669.  The  preceding  method  may  be  used  to  show  to  the 
eye  the  relation  between  two  varying  quantities  connected  by 
an  algebraic  equation.  The  following  is  an  example  of  the 
process.    Let  us  have  the  equation 

y  =  X  +  ^• 

We  suppose  x  to  hcve  in  succession  a  number  of  different 
values,  say  -4,-3,-2,-1,  0,  1,  2,  3,  and  4,  and  for 
each  of  these  values  we  calculate  the  corresponding  value  of  y. 

We  arrange  these  values  together,  thus: 

l^^^^^ofx -4,-3,-2,-1,     0,1,     2,3,    4. 

CoiTesp.valuesof2^..      6,       3i,     2,       li,  1,  Ji,  2,  3i,  5. 

We  next  draw  a  hori-  \ 
zontal  base  WX,  and  lay  \ 
the  values  of  x  off  on  it, 
the  positive  values  being 
laid  off  toward  the  right, 
the  negative  ones  toward 
the  left.  Then  from  each 
point  on  the  line  we  meas- 
ure upwards  a  length  equal 
to  the  corresponding  value  of  y,  and  there  make  a  point. 


TO", N.         -  .,  .. 

"     -4  -J  -I    -I 


274       BOOK  YU.    OF  LOOI  AND  OONIO  SEOTIONS. 


Let  the  points  be  a,  J,  c,  etc.  The  heights  of  these 
points  show  the  several  values  of  y.  Should  any  values  of  y 
be  negative,  they  extend  below  the  line.  Then  we  draw  a 
curve  through  all  the  points.  The  height  of  this  curve  above 
the  base-line  will  then  represent  the  value  of  y  corresponding 
to  any  value  of  x  between  the  limits,  -f  4  and  —  4,  so  that 
the  eye  can  see  at  a  glance  how  y  varies. 

670.  Dtf.  AbsoisBas  are  the  values  of  a?,  laid  off 
upon  the  base-line. 

571.  Def.  The  base-line  is  called  the  azlf  of 
abfloissas,  or  the  axis  of  X 

672.  Drf.  The  equation  which  gives  rise  to  a 
particular  curve  is  called  the  equation  of  that  curve. 

Exercises.  Plot  in  the  above  way  the  curves  correspond- 
ing to  the  following  equations  between  the  limits  a;  =  —  4 
and  a;  =  4. 

y  =  a;  -f-  4. 


2 


y  =  ~  _  3a;  -  1. 
3. 


y  =  —  -  3a: 


2 


these 

IS  of  y 
[raw  a 
above 
inding 
0  that 

id  off 
If  of 

to  a 
rve. 

ipond- 
=  -4 


GEOMETBY  OF  THEEE  DIMENSIONS. 


BOOK  VIII. 
OF  LINES  AND  PLANES. 


573.  Def   Geometry  of  three  dimensloM  treats 
HT^^  ^         ^^^^  ^""^  ''''*  ''''''^''^^  to  a  single 

.^f''''"'*7''f.i^'''  dimensions  is  also  called  the  ^.om^/ry 
of  space  and  solid  geotnetry.  ^  ^"Hn^ry 

^7^  ^/-   ^a>^el  Plaaeg  are  those  which  never 
meet,  how  far  soever  they  may  be  produced. 

6  T5.  Def.   A  straight  line  is  said  to  be  paraUel  to 

a  plane  when  it  never  meets  the  plane,  how  far  soever 
It  may  be  produced. 

676.  The  different  parts  of  a  figure  are  said  to  lie  in  one 
plane  when  a  plane  can  be  passed  through  them  alL 

Remark.  Whenever  a  plane  can  be  passed 
through  a  system  of  lines  or  points  in  space, 
the  theorems  of  plane  geometry  apply  to  the 
figure  formed  by  such  lines  and  points. 
Otherwise  they  do  not  so  apply, 

677.  Axiom  I.  If  two  or  more  points 
of  a  straight  line  lie  in  a  plane,  the  whole  Ime 
lies  m  that  same  plane. 

Ax.  II.  Any  number  of  planes  mav  be 
passed  through  the  same  straight  line,  and  a 
plane  may  be  turned  round  on  any  hne  lying 

Ax.  III.  Only  one  plane  can  pass  through     a,™, 
a  line  and  a  point  without  that  line  paS^^w^^l??^ 

samestiilghtli,;!^ 


276 


BOOK  VIII.    OF  LINES  AND  PLANES. 


EXERCISES. 

The  following  exercises  arc  inserted  here  in  order  that  the  student 
may.  by  working  upon  them,  acquire  a  clea\^«"««P\^°^  «'J?^"^f  ^''■ 
^nJe  between  the  relations  of  lines  in  space  and  in  a  plane  before  pro- 
ceeding to  the  study  of  the  theorems  of  the  geometry  of  space  If  four 
stTfl  rods  be  Jointed  together,  or  merely  held  together  so  as  to  form  a 
auadrUateral  it  may  assist  in  Exercises  8  and  4.  A  beginner  in  the 
subS^^  it  useful  to  construct  the  diagrams  in  space  with  wires, 

strines.  and  rods  attached  to  a  flat  board. 

Tpolygon  in  space  may  be  formed  by  joimng  end  to  end 
any  number  of  finite  straight  lines,  as  defined  in  Book  II., 
8142  The  only  change  is  that  in  Book  II.  the  hnes  are  all 
supposed  to  be  confined  to  one  plane,  whereas  there  is  no  such 
restriction  upon  polygons  in  space.    Now: 

1    Explain  that  all  the  propositions  of  Book  I.  whic) 
reto  to  triangles  are  true  of  triangles  in  spaxie,  however 

''^'^Thif  may  be  done  by  Theorem  I.,  which  follows,  by  showing  that  the 
three  sides  must  be  all  in  one  plane.  j.  .  t  «^i„„«„« 

a.  Explain  that  these  proportions  are  not  true  of  polygons 
of  four  or  more  sides  situated  anyhow  in  space. 

Show  that  a  polygon  of  four  or  more  sides  (which  may  be  formed  of 
stifl^traTgM  may  be  so  joined  that  its  sides  shall  not  all  be  m  one 

^^^""t  How  many  different  planes  may  the  sides  of  a  quadri- 
lateral in  space  contain  when  taken  two  and  two  ? 

4.  Show  that  the  two  diagonals  of  a  quadrilateral  in  space 

do  not  necessarily  intersect.                      ^  ,       a  «^„«^  ««  thA 
Show  that  each  pair  of  adjacent  sides  may  be  turned  round  on  the 
vertices  joining  tnem  to  the  opposite  pair.  

5  But  if  we  draw  three  lines  in  a  triangle,  one  from  each 
angle  to  any  point  of  the  opposite  side,  each  of  these  lines 
will  intersect  the  two  others. 

6  Any  pair  of  parallel  lines  must  lie  in  the  same  plane. 
But  there  may  be  three  lines  each  parallel  to  the  other  ^o, 
and  yet  they  may  not  all  three  lie  in  any  one  plane.  How 
many  planes  will  they  lie  in  when  taken  two  and  two? 

7.  If  four  lines  are  parallel,  how  many  planes  may  they 
lie  in  when  taken  two  at  a  time?  . 

8.  A  transversal  intersecting  a  pair  of  parallel  lines  lies  in 

the  same  plane  with  them. 


student 
3  dlfifer- 
>rc  pro- 
If  four 
form  a 
r  in  the 
li  wires, 

to  end 

lok  II., 

are  all 

ao  such 

which 
lowever 

;  that  the 
>olygon8 

formed  of 
be  in  one 

I  quadri- 
in  space 
nd  on  the 

rom  each 
lese  lines 

ae  plane, 
ther  two, 
le.    How 
►? 
may  they 

aes  lies  in 


RELATION  OF  LINES  TO  A  PLANE.  377 

th«til  ^r  ^'"""^  ,"°.*  '''  *^'  '^"^^  P^*^«  «^«  intersected  by 
tl  Z!  ?^"Tr^'  ^^  °^«"y  planes  may  be  determined  by 
me  three  hnes  taken  two  and  two?  ^ 


-» ♦  • 


CHAPTER   I. 

REUTION  OF  LINES  TO  A  PUNE 


Theorem  L 

578.    prot^^^  two  intersecting  straight  lines  me 
pcane,  and  one  only,  may  pass. 

in  ^F'ointo    ^""'^  '^'*'^^^  ^'''®''  "^^  ^^  ^^*  ^'n Meeting 

Conclusion.  One  plane,  and  A. 
only   one,   can    pass   through 
them. 

Proof.     1.    Let  any  plane 
pass  through  the  line  AB.  ^ 

2.  Turn  this  plane  around  on  AB  as  an  a-ria  «*,+,•!  •* 
any  point  C  of  the  line  CD  (§  577   II )  *  '"^*' 

conditions  (§  577  ifl)      Q  E  D  ^  ^  "''  *""^*'  *''* 

580.  Cor.  a.   TAroM^fA  anv  three,  nninu  «/.'  .V  «. , 

*««y«  «„.  onefhm,  andonlione,  n^pasT'      ""  """^ 


278 


BOOK  Yin.     OF  L1NB8  AND  PLANES. 


Theorem  II. 

681.    Tkoo  planes  intersect  each  other  in  a  straight 
line. 

Hypothesis.  MN,  PQ,  two  planes  intersecting  each  other 
along  the  line  AB. 

Conclusion.    The  line  ,     .  „ 

AB  18  a,  straight  line. 

Proof.  Let  A  and  B 
be  any  two  points  common 
to  both  planes. 

Draw  a  straight  line  M 
from  A  to  B.     Then — 

1.  Because  the  points 
A  and  B  are  both  in  the  plane  MN,  the  straight  line  AB  lies 
wholly  in  the  plane  MN  (§  677,  I.). 

2.  Because  the  points  A  and  B  are  both  in  the  plane  PQ, 
the  straight  line  AB  lies  wholly  in  the  plane  PQ. 

3.  Hence  this  straight  line  lies  in  bo+h  planes,  and  there- 
fore forms  their  line  of  intersection. 

4.  The  planes  cannot  intersect  in  any  point  not  in  this 
line,  because  then  two  planes  would  each  contain  a  line  and  a 
point  without  it,  which  is  impossible  (§  577,  III.). 

Therefore  ^^  is  the  only  line  of  intersection.     Q.E.D. 

683.  Corollary.  The  line  of  intersection  of  two  planes 
is  a  line  lying  wholly  in  both  planes. 

Theorem  III. 

583.  If  a  straight  line  he  perpendicular  to  two 
straight  lines  in  a  plane,  it  will  be  perpendicular  to 
every  other  straight  line  lying  in  the  plane  and  pass- 
ing through  its  foot. 

Hypothesis.  MN,  any  plane;  AB,  CD,  two  lines  in  this 
plane;  OP,  a  common  perpendicular  to  these  lines  at  their 
point  of  intersection  0;  BS,  any  other  line  lying  in  the  plane, 

__J wwi XT 1-        •^ 


a 


w 


th 

Tl 
wl 


BBLATtON  OP  LINES  TO  A  PLAm. 


M 


279 


Conclusion,     The  linA  np  ;«  „i 

Proof.     In  the  lines  '  Perpendicular  to  m 

^B  and  CZ>  take  the 
points  ^,  j5,  C  and  D, 
80  that 

AO^BO,    CO  =D0, 

Join  ^i)  and  A  C,  and 

let  i?  and  S  be  the  points 

m  which  the  joining  lines 
intersect  the  third  line 

R8,    (These  lines  must  interserf  J?Si  i.««         L^. 

-me  plane  with  it  and  notScf  to  it T"'*  ''^^  *"  "'  "'« 
Jom  PC,  PR,  pj,  p^  p  -^ 

Oc'^^nn   '^r^'"'  <'^^-<'  ^S?. 

tH.et.an«leat?laer,^:^^-f'-^^: 

BD  =  Aa 
Angle  (>^i?  =  angle  0^^. 
f^gJe  OC^  =  angle  OZ)^. 

toiLpSi^Tor;^'''''  *^«  equality  Of  ^O.i 

3.  Because  OP  is  perp^n,Uc„,ar  to  ^^.^  and  OA  =  OB. 
and,  in  the  same  way,      PC=  pn 

4.  Beoanse  of  these  equalities,  and  otBD=AO 

5.  Because,  m  the  triangles  PAR  and  PBS 

PA  =  P^, 
^i?  =  B8, 

f hooo  +  •      .        ^^^®  P^^  =  angle  P^^  • 
these  tnangles  are  identically  equalf  and       ' 

pT>  -pa 

Therefore  ^gle  POsZ^x^gh  pqs- 

whence  POifa^dPO^are  right  ai:,e?^',  OP  .    ^c 
"•  "^"■'"'"  ^'^'  ""•?  •«  »-y  '^e-whateveriyin^gf-the 


(3) 
(2) 
W 


280 


BOOK  VIIL     OF  LINES  AND  PLANES. 


plane  and  passing  through  0,  the  line  OP  is  perpendicular 
to  every  such  line.     Q.E.D. 

584.  Def.  A  line  meeting  a  plane  so  as  to  be  per- 
pendicular to  every  line  lying  in  the  plane  and  pass- 
ing through  the  point  of  intersection  is  said  to  be 
perpendicular  to  the  plane. 

586.  Corollary,  From  a  given  point  in  a  plane  only  one 
perpendicular  to  the  plane  can  be  erected,  and  from  a  point 
without  a  plane  only  one  perpendicular  can  be  dropped  upon 

the  plane. 

Theorem  IV. 

586.  Conversely,  all  lines  perpendicular  to  an- 
other line  at  the  same  point  lie  in  the  same  plane. 

Hypothesis.     OP,  any  straight  line ;  0-4,  OC,  two  lines 
perpendicular  to  OP  at  0; 
OB,  any  third  line  per- 
pendicular to  OP  at  0. 

Conclusion.  OB  lies 
in  the  same  plane  with 
OA  and  OC. 

Proof.     1.    If   OB  is 
not  in  the  plane  A  00, 
pass  a  plane  through  PO  and  OB,  and  let  OB'  be  the  line  in 
which  it  intersects  the  plane  AOC. 

2.  Because  OP  is  perpendicular  both  to  OA  and  OC,  it  is 
also  perpendicular  to  OB',  which  lies  in  this  plane  (§  583). 

3.  Because  OB'  is  in  the  plane  POB,  we  have  in  this 
plane  two  straight  lines,  OB  and  OB',  both  perpendicular  to 
OP,  which  is  impossible. 

4.  Therefore  OB  and  OB'  are  the  same  straight  line,  and 
OB  lies  in  the  plane  A  OC.    Q.E.D. 

587.  Corollary.  If  a  right  angle  be  turned  round  one  of 
its  sides  as  an  axis,  the  other  side  will  describe  a  plane. 

Theorem  V. 
>    688.  ^  a  plane  bisect  a  line  perpendicularly, 
every  point  of  ihe  plane  is  equally  distant  from  the 
ends  of  the  line. 


BBLATlOif  OF  Lims  TO  A  PIANB.  281 

intone  tt  afl"  ""  '""'^'""'"'"  '^  "">  »"-  ^^. 
middle  point  0  of  the  |P 

line ;  i?,  any  point  in 
the  plane. 

Conclusion,  R  ig 
equally  distant  from  P 
and  Q. 

Proof.    Join    OR, 
Then— 

1.  Because   P^  is 
perpendicular    to    the  '0 

plane,  it  is  perpendicular  to  OR  in  that  plane  (§  684). 

2    Because  OR  is  perpendicular  to  PQ  at  its  middle  point 
0,  It  18  equally  distent  from  P  and  Q  (§  104).    ^i  D 

589.  Cbro//«ry.  Conversely,  mr^^ ;>om^  ^A,-^;,  •,  ,'„ 
distant  from  two  fixed  points  is  in  the  plane  bisecting  TriZ 
angles  the  line  joining  the  points,  ^        ^ 

Theorem  VI. 

ir/^rtUrP^^'  *'"  I«!^"^-'-  to  a  plane 
and  Q. 

Conclusion.  These 
perpendiculars  are  par- 
allel. 

Proof.    In  order  to 
proye   the    parallelism  « 
of  the  lines,  we  must 
show — 

I.  That  they  can  never  meet; 

T   Tf+i,  ?■  T?"**  *^®y  ^^e  in  the  same  plane. 


-    ^-;-  ■, 


'.1     !■ 


282 


BOOK  VIIL    OF  LINES  AND  PLANES. 


II.  From  the  point  P  draw  in  the  plane  JlfiV  the  line  PAy 
perpendicular  both  to  PQ  and  to  PR.  In  the  line  Q8  take 
QB  ^  PA.    Join  BP,  BA,  QA.    Then— 

1.  Because,  in  the  right-angled  triangles  QPA  anl  PQB, 

AP  =  BQy  PQ  common, 
these  triangles  are  identically  equal,  and 

BP  =  AQ. 

2.  Because,  in  the  triangles  BQA  and  BPA^ 

AQ  =  BP,  BQ  =  AP,  AB  common, 
these  triangles  are  idexitically  equal.     But  BQA  is  a  right 
angle  (§  584).     Therefore  tue  corresponding  angle  APB  is 
also  a  right  angle. 

3.  Therefore  from  the  point  P  of  the  line  AP  there  pro- 
ceed tLiree  straight  lines  PQ,  PB,  and  PR,  all  at  right  angles 
to  AP.  Hence  these  tlnee  lines  are  in  one  plane;  that  is, 
PR  is  in  the'  plane  fixed  by  the  two  lines  PQ,  PB, 

4.  But  QS  is  also  In  this  plane,  because  it  intersects  these 
lines  (§  579).     Therefore  QS  and  PR  are  in  the  same  plane. 

5.  Kence  the  lines  PR  and  QS  are  in  the  same  plane  and 
never  meet,  and  are  therefore  parallel.    Q.E.D. 

Theorem  VII. 

591.  Conversely,  if  one  of  several  parallels  is 
porpendicv^ar  to  a  plane,  each  of  the  others  is  also 
perpendicular  to  tftat  plane. 

Hypothesis.    A  plaiie,  MN',  two  parallel  lines,  PR  and 
QS,  intersecting  the  plane  at  P 
and  Q  in  such  manner  that  QS 
is  perpendicular  to  the  plane. 

Conclusion.  Pi2  is  also  per- 
pendicular to  the  plane.  Jf^ 

Proof.  1.  n  PR  is  not 
perpendicular  to  the  plane,  let 
PR'  be  perpendicular  to  it.         /' 

2.  Then  PR'  is  parallel  to  ^ 

QS  (§  590). 

3.  Therefore'  through  the  point  P  we  hare  two  straight 
lines,  PR  and  PR',  both  parallel  to  QS,  which  is  impossible. 


BT. 


B 


7" 


h 


I 


RELATION  OF  MNES  TO  A  PLANS.  ggs 

4.  Therefore  the  perpendicular  to  the  plane  at  Pis  the 
line  Pi?.    Q.E.D 

Corollary.  It  two  or  more  lines  are  each  parallel  to  the 
same  straight  line,  and  a  plane  be  drawn  perpendicular  to 
this  straight  line,  it  will  also  be  perpendicular  to  the  other 
lines,  and  they  wiU  be  parallel  to  each  other.    Hence: 

593,  Lines  parallel  to  the  same  straight  line  are  paral- 
lel to  each  other. 


V 


Theoeem  VIII. 

693.  Frora  any  point  above  a  plane  lines  Tneeting 
the  plane  at  equal  distances  from  the  foot  of  the  per- 
pendicular are  equal,  and  the  line  meeting  the  plane 
at  the  greater  distance  from  this  foot  is  the  greater. 

Hypothesis.  MN,  a  plane;  P,  any  point  outside  of  it; 
Oy  the  foot  of  the  per- 
pendicular from  P; 
A  OG,  any  straight  line 
in  the  plane  through 
O'j  Ay  B,  two  points  in 
the  plane  equally  dis- 
tant from  0;  G,  a  point 
more  distant  from  0 
than  A  is. 

Conclusions.     I.  PB  =  PA. 
II.  PC  >  PA. 
Proof.    1.  In  the  triangles  POA  and  POB, 

PO  is  common. 
OA  =  OB  (hyp.). 
Angle  POA  =  angle  POB  (both  right  angles). 
Therefore  hypothenuse  PB  =  hypothenuse  PA.     Q.E.D. 

2   Because  0  is  the  foot  of  the  perpendicular  from  P  on 
AC,  and  0C>  OA, 

BC>PA{^10B).     Q.E.D. 
„„^.   .-^,^,,^,y  ^.     ^j   i,nrougn  ihe  centre  of  a  circle  a 
line  he  passed  perpendicular  to  its  plane,  each  point  of  this 
Aue  IS  equally  distant  from  all  points  of  the  circle. 


acz= 


284 


BOOK  VIIL    OF  LIITES  AND  PLANES. 


'  595.  Cor.  2.  Equal  lines  meet  the  plane  at  equal  dis- 
tances from  the  foot  of  the  perpendicular,  and  greater  lines  at 
greater  distances.  ^ 

This  corollary  may  be  expressed  as  follows: 

596.  The  locus  of  the  point  in  a  plane  at  a  given  dis- 
tance from  a  fixed  point  without  the  plane  is  a  circle  drawn 
around  the  foot  of  the  perpendicular  as  a  centre. 

597.  Def.  The  pr<\jection  of  a  point  upon  a  plane 
is  the  foot  of  the  perpendicular  dropped  from  the 
point  upon  the  plane. 

Example.  If  MN  be  a  plane,  and  P  a  point  outside  of 
it,  and  if  the  perpendicular  from  P  upon  the  plane  meet  the 
latter  in  P',  then  P'  is  the  projection  of  P  upon  the  plane 

Mm 

598.  Def.  Tlie  projection  of  a  line  upon  a  plane 
is  the  locus  of  the  feet  of  the  perpendiculars  dropped 
from  every  point  of  the  line  upon  the  plane. 

Example.  The  line  P'R'  is  the  projection  of  the  line 
PR  upon  the  plane  ifiV". 


Theorem  IX. 

599.  The  projection  of  a  straight  line  upon  a 
plane  is  itself  a  straight  line,  and  the  straight  line 
and  its  projection  are  in  one  plane. 

Hypothesis.  ifJV,aplane;  P^i?,  a  straight  line;  P'Q'R', 
the  projection  of  PQR 
upon  JOT. 

Conclusion.  P'Q'R* 
is  a  straight  line,  and  lies 
in  one  plane  with  PQ. 

Proof.  1.  Because  the 

lines  PP'  and  RR'  are  

perpendiculars    to      the  ^N 

plane  MN^  they  are  parallel  to  each  other,  and  therefore  in 
one  plane  (§  690). 


BBLATION  OF  LINES  TO  A  PLANE.  335 

2.  Pass  a  plane  through  these  lines.    Because  this  plane 
contams  the  points  P  and  R,  it  will  contain  the  point  0 
which  lies  on  the  Kne  PR.  ^         ^' 

3.  Because  the  line  QQ'  is  perpendicular  to  MN,  it  is  also 
parallel  to  PP^(%m);  and  because  it  contains  'the  pott 
Q,  the  plane  PP'QQ'  is  the  same  as  the  plane  PP'Rr' 
Therefore  the  foot  Q'  lies  in  this  same  plane.        '  ^^  ^^  ' 

T    ^;fi^o?wl  *^®  intersection  of  two  planes  is  a  straight 
me  (§  681),  the  foot  Q',  which  lies  on  the  intersection  of  the 
two  planes,  is  m  a  straight  line  with  P'  and  R' 

5.  Because  ^  may  be  any  point  on  PR,  the*  projection  of 
every  point  of  PR  is  in  the  straight  line  P'R\    Q.E.D 

/,w^^^'  ^,V^\y  ^'     Va  line  intersect  a  plane,  its  projec^ 
tion  passes  through  the  point  of  intersection.  ^^ 

601.  Cor.  2.  If  aline  le  perpendicular  to  a  plane  its 
pro^ectnn  upon  the  plane  is  a  point;  namely,  thepTnt  n 
which  %t  intersects  the  plane. 


Theoeem  X. 

ar,nJ^^.nJ{^,  ^m^  mjJ^^^ciJ  a  plane,  it  makes  a  less 
angle  with  its  projection  than  with  any  other  line  in 
the  plane  passing  through  the  point  ofintersecZi 

vi.S'aroAS;^^^^^^^  "^'  ^  ^^^^  ^^^--^^^^  *^^« 

jection  of  OD  upon  the 
plane;  0^  any  other  line 
m  the  plane  passing 
through  0. 

Conclusion.  The  angle 
^OA  is  less  than  BOB. 

Proof.  Take  OB  = 
OA,  and  join  AB 
and  DB.    Then— 

of  ij,  ^'""^'^  ^^^  ''  ^  ^^''  '^^gl«  (^  being  the  projection 


DB  >  DA. 


(§593) 


d8d 


BOOK  VITL    OF  LINES  AND  PLANES. 


2.  Because  the  triangles  DO  A  and  DOB  have  the  side  DO 
common,  and  OB  =  OA,  while  the  third  side  DB  is  greater 
than  the  third  side  DA, 

Angle  DOB  >  DOA  (§  116).    Q.E.D. 

603.  Def.  The  angle  between  a  line  and  its  pro- 
jection on  a  plane  is  called  the  inollnation  of  the  line 
to  the  plane. 


l: 


1! 


Theorem  XI. 

604.  ^  a  line  intersect  a  plane — 
I.  The  angle  which  it  makes  with  a  line  in  the 
plane  passing  through  its  point  of  intersection  is 
greater,  the  greater  the  angle  this  last  line  makes 
with  its  pfojection. 

II.  The  line  makes  equal  angles  with  lines  at  equal 
angles  on  both  sides  of  its  projection. 

Hypothesis.  MN,  a  plane;  OD,  a  line  intersecting  it  in 
O'y  OA,  the  projec- 
tion of  OD  on  the 
plane;  OB,  OB', 
two  lines  from  0 
making  equal  angles 
with  OA;  00,  a 
line  making  a  still 
greater  angle  with 
OA. 

Conclusions.      I.    Angle  DOC  >  angle  DOB. 
II.    Angle  DOB  =  angle  DOB\ 

Proof  I.  Take  OB,  OB',  and  0C7aU  equal  to  OA.  Join 
DB',  DA,  DB,  DC,  AB,  AB',  BC.    Then— 

1.  Because  the  points  B,  A,  B',  C  are  all  in  the  same 
plane  and  equally  distant  from  0,  they  lie  on  a  circle  having 
0  as  its  centre. 

3.  Because  angle  AOC>  A  OB,  the  distance  ACia  greater 


XI it-  -    -i J     J  r»- 

iiuan  lui)  uuoiu  ^x>; 


XT *— ~ 


DC>DB. 


(§  693) 


RELATION  OF  LINES  TO  A  PLANE. 


287 


3.  In  the  ti-ianglos  ODB  and  ODC  we  have 

0/?  common;  0C=  OBy  DC>  DB, 
Therefore  ? 

Angle  DOG  (opp.  DC)  >  angle  DOB  (opp.  DB),     (§  115) 

Q.E.D. 

Proof  II.    4.  In  the  triangles  A  OB  and  u4  OB'  we  have 
0^  common; 

OB  =  0-ff'  (construction); 
Angle  AOB  =  angle  ^ 05'  (hyp. ). 
Therefore  these  triangles  are  identically  equal,  and 

AB  =  AB\ 
6.  Because  DAB  and  DAB'  are  both  right-angled  at  A 
and  because  /?^  is  common  and  AB  =  ^i5',  these  trianries 
DAB  and  jD^jB'  are  identically  equal,  and 

DB  =  DB'. 
6.  Therefore  the  triangles  DOB  and  J905',  having  their 
sides  equal,  are  also  identically  equal,  and 

Angle  Z>05  =  angle  Z?(?5'.    Q.E.D. 


Join 


Theorem  XII. 

606.  At  the  point  of  intersection,  a  line  in  the 
plane  perpendicular  to  the  projection  of  a  line  is 
perpendicular  to  the  line  itself. 

Hypothesis.    OD,  a  line  intersecting  the  plane  in  0-  OA 
the    projection    ot    OD  or  ,     ^, 

upon  the  plane;  POQy  a 
line  in  the  plane  perpen- 
dicular to  OA. 

Conclusion. 
Line  POQ  ±  OD. 

Proof.        Take     the 
points  P  and  Q  at  equal  distanc  as  from  0,  and  join  AP,  AQ, 

1.  Because  the  points  P  and  Q  are  at  equal  distances  from 
biiv  xv^uu  {y  ui  mc  purpuudicujar  ^  c/  on  PQ,  we  have 


!288 


BOOK  VIIL    OF  LINES  AND  PLANES. 


2.  In  the  trianglea  DAP  and  DAQ, 

DA  is  common; 

AP=^AQ',  s       (1) 

Angle  DAP  =  angle  DAQ  (hyp.). 
Therefore  DP  =  PQ. 

3.  In  the  triangles /)0i^  and  i>Og, 

i>0  is  common; 
DP  =  DQ;  (2) 

OP  =  0^  (construction). 
Therefore  Angle  DOP  =  angle  DOQ; 

and  because  PO^  is  a  straight  line,  both  these  angles  are 
right  angles.     Q.E.D. 

Corollary.  If  the  line  OD  is  not  perpendicular  to  the 
plane  MN,  there  can  be  only  one  line  in  this  plane  perpen- 
dicular to  OD.  For  if  there  were  two  such  lines,  OD  would 
be  perpendicular  to  the  plane  (§  583).  Hence,  because  POQ 
is  the  only  perpendicular  to  OD  in  the  plane : 

606.  Conversely,  a  line  in  a  plane  perpendicular  to  an 
intersecting  line  is  perpendicular  to  the  projection  of  the  inter- 
secting line. 

Sdwlium.  An  astronomical  illustration  of  these  theorems  is  afforded 
by  conceiving  one's  self  to  be  looking  at  the  sun  in  the  south.  The 
plane  is  that  of  the  horizon,  in  which  the  observer  must  suppose  himself 
situated  at  the  point  0.  Let  the  line  OD  be  that  toward  the  sun.  (It  is 
not  necessary  to  suppose  it  cut  off  at  D  or  any  other  point,  because  our 
theorems  do  not  refer  to  its  length.) 

Then  the  horizontal  line  OA  from  the  observer  to  that  point  of  the 
horizon  under  the  sun  will  be  the  projection  of  the  line  to  the  sun. 

By  Theorem  X.  the  angular  distance  of  the  sun  from  this  point  will 
be  less  than  from  any  other  point  of  the  horizon.  This  angle  is  called 
the  sun's  altitude. 

If  we  suppose  a  horizontal  east  and  west  line,  Theorem  XII.  shows 
that  this  line  will  always  be  at  right  angles  both  to  the  direction  of  the 
sun  and  to  the  south  line  which  passes  directly  below  the  sun. 

If  we  take  a  series  of  points  along  the  horizon,  starting  from  the 
point  directly  below  the  sun,  Theorem  XI.  shows  that  the  angular  dis- 
tance of  these  points  from  the  sun  will  increase  up  to  the  opposite 
point  of  the  horizon  from  that  below  the  sun. 


BBLATION  OF  LINES  TO  A  PLAXB. 


289 


(1) 


(8) 


are 


the 


Theorem  XIII. 

607.  When  two  straight  lines  are  parallel,  each 
of  them  is  parallel  to  e&erp  plane  passing  through 
the  other  and  not  containing  both  lines. 

Hypothesis,  AB,  CD,  two  paraUel  straight  lines;  MI^, 
any  plane  passing  through  CD, 

Conclusion.     AB  ia     a.- 
parallel    to    the  plane 

JfiV. 

Proof.  Let  us  call 
P*  the  common  plane 
of  the  parallels  AB  and 
CD.    Then— 


~B 


1.  Because  AB  lies  wholly  in  the  plane  P,  if  AB  meets 
the  plane  Mli  at  any  point,  that  point  will  be  common  to 
the  plane  P  and  the  plane  MN. 

2.  But  the  only  points  common  to  these  two  planes  are 
on  their  line  of  intersection;  namely,  the  line  CD  (§  682) 
Therefore  if  AB  eyer  meets  the  plane  JfJV,  it  must  meet  this 
line  CD. 

3.  But,  by  hypothesis,  it  is  parallel  to  CD,  and  so  cannot 
meet  it. 

4.  Therefore  it  cannot  meet  the  plane  JO^,  and  therefore 
IS  parallel  to  it  (§  575).     Q.E.D. 

5.  But  should  the  plane  JfiV^  coincide  with  the  plane  P, 
the  line  AB  will  then  lie  in  JfiV^as  it  does  in  P. 

Theorem  XIV. 

608.  Angles  of  which  the  sides  are  parallel  and 
similarly  directed  are  equal. 

Hypothesis.    BOC  and  P'O'C",  two   angles   in   which 

OPilO'P'and  0C\\  O'C, 
Conclusion.    Angle  BOC—  angle  B* 0' C\ 

*  The  letter  Pis  here  employed  not  as  a  mark  on  the  diagram,  but 
as  a  short  and  convenient  appellation  of  the  plane  referred  to.  Such  a 
use  of  letters  is  of  constant  occurrence  in  the  higher  geometrv  and 
should  be  well  understood.  ^  ^' 


'^90 


BOOK  VIIL    OP  LINES  AND  PLANES. 


Proof.     On  the  sides  B  and  B'  take  OB  =  0'B\  and 
on  the  sides  C  and  C  take 
OC  =  0'C\      Join   BB', 
CC\  00',   Then— 

1.  Because  05  and  0'^' 
are  equal  and  parallel,  the  q^ 
figure  OO'BB'  is  a  parallel- 
ogram  (§  138),  and 

BB'  =  and  ||  00\ 

2.  In  the  same  way, 
CC:  =  and  ||  00\ 

3.  Therefore  BB'CQ'  is  a  parallelogram,  and  BC 

4.  Therefore  the  triangles  i^OC'and  B'O'C,  having  the 
three  sides  of  the  one  equal  to  the  sides  of  the  other,  are  iden- 
tically equal,  and 

Angle  BOC  =  angle  B'0'C\    Q.E.D. 

609.  Corollary.  It  may  be  shown,  as  in  Book  II.,  Th. 
VI.,  that  if  the  sides  of  the  angles  are  parallel  and  oppositely 
directed,  the  angles  will  still  be  equal,  and  that  if  one  pair  of 
sides  is  similarly  directed  and  the  other  oppositely  directed, 
the  angles  will  be  supplementary. 

Theorem  XV. 

610.  Parallel  lines  intersecting  the  same  plane 
make  equal  angles  with  it. 

Hypothesis.     OA,  PB,  two  parallel  lines  intersecting  the 
plane  MN  in   0  and  P; 
OA',  PB'y  the  projections 
of  certain  portions  of  these 
lines  on  the  plane. 

Conclusion. 
Angle  A  0^'=:angle5P5'. 

Proof.    At  the  points 
0  and  P  erect  the  per- 
pendiculars OR  and  P8. 
Then— 

1.  Because  OR  and  PS  are  perpendicular  to  the  same 
plane,  they  are  parallel  (§  590),  while  OA  \\  PB^  by  hypothesis. 
Hence  Angle  AOR  =  angle  BPS.  (§  608) 


and 


BBLA  TION  OF  UNES  TO  A  PLANS,  ggj 

an/'^^T'"  ^^i'  *  traneversal  crossing  the  parallels  A' A 
and  OR  It  IS  m  the  same  plane  with  them.  Also  becau^ 
OR  ±  plane if^;  OR  1  OA'  m  that  plane  (§584^  The^m^ 
ehmgs  are  true  of  Pi?',  PB,  and  PS. 

3.  Because  A' OR  and  B'PS  are  right  angles.  ^  0^'  is  thfi 
complement  of  the  angle  A  OR,  and  BPB^  it  t^e  compleme^ 
of  the  equal  angle  BPS.    Comparing  with  (1),     '''''^^^^'^'"'^ 
Angle  A  OA'  =  angle  BPB\     Q.E.D. 

Theoeem  XVI. 

n^^^\'  ^^^^^^  ^^^  ^^'^^^  ^ot  in  the  same  plane  me 
and  only  one,  comTnon  perpendicular  can  hed^raZ 
Hypothesis.    AB,  CD,  two  lines  not  in  the  same  plane 
and  therefore  passing  each  other  without  intersecting 


to  bi"^rn/^r ''  ^"  ^"^^'"^  ^^  -°^^'  p-p-^-^- 

Ipf  ,>  r"-^'  ^'  T^'"''''^^  '^'''  ^^^^'  «^y  ^A  pass  a  plane,  and 
let  t  turn  round  on  OD  until  it  is  parallel  to  AB.  Let  ^^ 
be  this  plane     Let  A'B'  be  the  projection  of  AB  on  the 

1.  Every  point  of  A  'B'  is  fixed  by  dropping  a  pernendion 
ar  from  some  point  of  AB.     Let  p\,  the^ofn'  ofThicr^ 
IS  the  projection.     Then  PO  1  plane  MN  (§  597) 

lar  t;  bofbT  f  ^  ^^P;^r^li«"I^r  to  JfiV,  it  is  perpendicu- 
lar to  both  the  lines  A'B'  and  CD,  which  lie  in  MN. 

o.  Hecanae  PO  is  nArnGnri]V.«]o^  f^  .«/n/  .^  • 
pendicular  to  AB,  which  is  parall      3  ^'5'  (8  72 

Therefore  OP  is  perpendicular  to  both  t 


OD.     Q.E.D. 


lines  AB  and 


292 


BOOKVIIL    OF  LINES  AND  PLANEa. 


II.  If  there  is  any  other  common  perpendicular,  let  it  be 
P'Q,   Through  ^  draw,  in  the  plane  JfiV,^i?lMi?.   Then— 

4.  Because  P'Q  is  perpendicular  to  AB,  it  is  also  perpen- 
dicular to  QR,  which  is  parallel  to  AB. 

6.  Because  P'Q  is  perpendicular  to  both  the  lines  QE  and 
CD,  it  is  perpendicular  to  their  plane  JfiVi 


6.  Put,  because  A'B'  is  the  projection  of  AB,  the  foot  of 
the  perpendicular  from  P'  on  the  plane  must  fall  on  some 
point  of  A'B'.     Let  0'  be  this  point. 

Therefore  from  the  point  P'  are  dropped  two  perpendiculars 
P'O'  and  P'Q  upon  plane  ifiV",  which  is  impossible  (§585). 

Therefore  P'Q  is  not  a  common  perpendicular  to  the 
lines  AB  and  CD,  and  PO  is  the  only  common  perpendicular. 

Q.E.D. 
Theorem  XYII. 

613.  The  least  distance  between  two  indefinite 
lines  which  do  not  intersect  each  other  is  their  corn- 
mon  perpendicular. 

Hypothesis,   a,  h,  two  lines  in  space,  the  one  being  sup- 
posed to  lie  behind  the  other, 
so  that  they  do  not  intersect. 

Conclusion.  No  line  which 
is  not  perpendicular  to  both 
lines  can  be  the  shortest  line 
between  them.  "-^  P        b 

Proof.  If  possible,  suppose  that  some  line  PQ  which  does 
not  make  a  right  angle  with  a  is  the  shortest  line. 

From  P  drop  a  perpendicular  PR  upon  a.     Then 

PR  <  PQ, 
Therefore  PQ  is  not  the  shortest  line. 


JtXLATIOJfS  OF  PLANBa. 


293 


Shortest  """^  *'  """^  ^°  ""^  ^^'^^^ '»'"« Ji""  ■»"»'  be  the 
Mn^XZy^  "°'  "  "'"'  ""*'"«  "  "«•"  '»«'«  »'"»  both 


i>>» 


® 


CHAPTER    II. 

REUTIONS  OF  TWO  OR  MORE  PUNES. 


Theorem  XVIII. 

inauL  w*^ ^"'''*  <^re parallel  ^any  two  intersect- 
tng  lines  on  the  one  are  both  parallel  to  the  other  plane 

parallel  to  the  plane  PQ. 

Conclusion,     The  planes 
i/"i\randP^  are  parallel. 

J'roof,  1.  If  the  planes 
are  not  parallel,  they  must 
intersect  in  a  straight  line 
lying  in  both  planes,  and 
therefore  in  the  plane  MM 
Let  us  call  this  line  X  * ^ 

^Aan'rwtr 'c:t?he  5::  7o'  "^^  "^  f '  r.  ^"^  '>"'- 

are  paraUel.    Q.E.D  '^  ^"   ^''««*°"^  *hese  planes 

intVstt  ImliV  '"^  '''^'^^   ^^-o  P<^ranel  pUnes 
traersect  a  third  plane  are  parallel. 


294 


BOOK  VIIL    OF  LINES  AND  PLANES. 


Let  ns  call  A  and  B  the  parallel  planes,*  and  Xthe  third 
or  intersecting  plane.     Then — 

1.  The  lines  of  intersection  are  in  one  plane,  because  they 
both  lie  in  the  plane  X. 

2.  Because  one  of  the  lines  of  intersection  lies  in  the 
plane  A,  and  the  other  in  the  plane  By  parallel  to  it,  and 
because  these  planes  never  meet,  the  lines  can  never  meet. 

Therefore  the  lines  are  in  one  plane  and  can  never  meet, 
and  so  are  parallel,  by  definition.     Q.E.D.  , 

Theorem  XX. 

616.  Parallel  planes  intercept  equal  segments  of 
parallel  lines.  '  • 

Hypothesis.  MN,  PQ,  two  parallel  planes;  AB,  CD,  two 
parallel  lines  intersecting  the  planes  in  the  points  A,  B, 
C,D. 

Conclusion.  AB  =  CD. 

Proof.  1.  Join  ^  C  and 
BD.  Consider  the  plane 
containing  the  parallels 
AB  and  CD.  Because  the 
four  points  A,  B,  C,  and 
D  all  lie  in  this  plane,  the 
joining  lines  A  C  and  BD 
lie  in  it. 

2.  But  because  the  lines  AC  and  BD  also  lie  in  the 
respective  planes  MNand  PQ,  they  are  the  lines  of  intersec- 
tion of  these  planes  with  the  plane  A  BCD. 

3.  Because  the  planes  ifiVand  PQ  sre  parallel  (hyp.).  ILo 
lines  of  intersection  A  C  and  BD  are  parallel  (§  614). 

4.  Because  AB  ||  CD  (hyp.)  and  AC  ||  BD,  as  just 
shown,  A  BCD  is  t*  parallelogram.     Therefore 

Ali-  (7i>(§127).     Q.E.D. 


*  This  theorem  iti  so  iiimple  that  the  student  can  imagine  the  figure 
which  is  to  embody  the  hjrpothesis  and  conclusion  better  than  it  can  be 
j'ork'ppaonto/i  in  a  dlaTam.  We  therefore  ffive-  the  demonstration  with- 
out a  diagram. 


2 
i 

1 


RELATIONS  OF  PLANES, 


905 


Theorem   XXI, 

616.  Planes  perpendicular  to  the  same  straight 
line  are  parallel  or  coincident.  ^ 

Hypothesis.    Two  planes,  MN  and  PQi  OH,  a  line  ner- 
pendicular  to  each  of  these  planes.  >     ^«  per 

Conclusion.     The  planes  are 
parallel 

Proof.  If  they  are  not  parallel, 
tKey  must  intersect.  If  they  in- 
tersect, call  Xany  point  on  the 
line  of  intersection  and  join  OX, 
RX,    Then- 

1.  Because  OX  is  in  the  plane 

S.''  "  P^^P^"^^^"!^^  *«  OR,  a  perpendicular  line  to  the 

to  0k^'°'"''  ^^''  '"^  *^'  ^^'"'  ^^'  '^  ''  ^^««  perpendicular 

3.  Therefore  from  the  same  point,  X.  we  have  two  r.«r 

rrs^/' ""'  ^''' '"  ''^  ^'  «t-4hTiSrwS 

4.  Therefore  the  planes  never  meet,  and  so  are  parallel. 

hJi^J'i  ^'"■""''7-   Conversely,  a  straight  line  perpendictl 
Ur  to  aplam  ts  also perpendioular  to  every  paralhl plane. 


Theokem  XXII. 
*«*;  f  ^*™^S^  ^«««  ■m<ikes  equal  angles 


parallel  planes. 

Hypothesis.  MN,  PQ, 
two  parallel  planes;  AB, 
a  straight  line  intersect- 
ing these  planes  at  the 
points  B  and  F;  EA\ 
FA*',  the  projections  of 
this  lino  upon  the  re- /' 
spective  planes. 


with 


m 


296  BOOK  VIII.     OF  LINES  AND  PLANES. 

Conclusion.  Angle  A EA '—that  is  (§  d03),  the  inclination 
of  AE  to  the  plane  MN-Ab  equal  to  angle  AFA",  the  incli- 
nation of  the  line  to  the  plane  PQ. 

Proof.  1.  Because  the  points  A'  and  A"  are  the  projec- 
tions of  the  point  ^-upon  parallel  pbnes,  the  point  A*  ^3  in 
the  straight  line  AA"  (§§  597,  «17). 

2.  Because  the  plane  of  the  two  linos  A  A''  and  'AE  con- 
tains the  four  points  A',  E,  A",  anc'  F,  the  lines  A*E  and 
A' 'Fare  in  this  same  plane,  and  are  its  line  of  intersecMon 
with  the  parallel  planes  M^F  and  PQ,     Therefore 

A'E  II  A"F  (§  614) 

and     Angle  ^-E^^' =  cor.  angle  ^i^^".    Q.E.D. 

Theorem  XXIII. 

619.  If  from  any  points  of  the  line  of  intersec- 
tion of  tw6  planes  two  perpendiculars  to  that  line  he 
drawn^  one  in  each  plane.,  they  will  form  equal  angles. 

Hypothesis.  M  and  iV,  two  planes  intersecting  along  the 
line  AB ;  OQ,  08,  two 
lines,  one  in  each  plane, 
perpendicular  to  AB', 
PR,  PT,  two  other  lines, 
one  m  each  plane,  per-  ^ 
pendicular  to  AB. 

Conclusion. 
Angle  QOS=  angle  RPT, 

Proof.  1.  Because  the 
sides  OQ,  PR  lie  in  the  same  plane  M,  and  are  perpendicular 
to  the  same  straight  line  AB, 

OQ  II  PR,  (§  70) 

2.  In  the  same  way, 

OS  II  PT. 

3.  Therefore 

Angle  Q08  =  angle  RPT (§  608).     Q.E.D. 

630.  Def.  Two  planes  which  intersect  are  said  to 
^nrm  a  Hihftilral  anerle  flloner  their  line  of  meetinsr. 

An  angle  formed  by  two  lines  is  called  a  plane 
angle  to  distinguish  it  from  a  dihedral  angle. 


i 


^!i 


a'. 


RELATIONS  OF  PLANES.  2Q7 

t 

i^^^V^'  ^"-C- .  "^^  '*•»"'  o'  a  dihedral  angle  are  fhn 
two  planes  which  fonn  it.  ^  '"® 

linf  ff  ■  ^f  ■  "^f  f  ^^  "'  ^  ^liedral  angle  is  the 
Ime  of  meeting  of  the  planes  which  fonn  it 

633.  A  dihedral  angle  is  measured  by  the  anrfn 
a^^^t^iredr^^''^-'  "''^  ^  eachVc:,S 

by  either  X  a^  go'/o:  fc  ^^  "^/^  "^'^"'^''^ 
diculars.  V^o  or  i,;i-/  between  the  perpen- 

which  side«  will  be  trLte?sf^tlw^f/i*^'"S^««*' 
^  (§  584).  Heuee  a  dihedral ^SV^ltti'-TA  ^^^ 
angk  between  the  intersections  of  itrfaceS  a'„i  ^''"" 
pendicniar  to  its  edge.  ™  *  P'^*  Per- 

dihetla^gL*™  todefinirintersecting  planes  form  fonr 

angL^thonW  JtronftS""'""',  ^^^""^  ^^^^'^^ 
from  the  oorrespontog  S^^ll  ^'^^  ^r  *■>«  student 
angles.  ^  Propositions  respecting  ordinaryplane 

TT    /7T1      T         ,     -^  ^niersemon  as  a  commnft  firing 
---.  ^^^v,o«.-  ,«,-«am  ««^fes  are  equal.      " 


I 


■If 


if- 


mtmi^m 


298 


BOOR  Vlll.     OF  LINES  AND  PLANES. 


Corollary.  When  two  planes  intersect,  we  may  take  either 
of  the  two  adjacent  dihedral  angles  as  measuring  their  incli- 
nation. 

Y.Ifa  plane  intersect  two  parallel  planes,  the  alternate 
and  corresponding  dihedral  angles  are  equal. 

Note.  When  speaking  of  the  angle  between  two  planes,  the 
adjective  dihedral  may  be  omitted  when  no  ambiguity  will  arise  from 
the  omission. 

Theoeem  XXIV.  \ 

635.  Iffrmt  anypoint  perpendiculars  he  dropped 
upon  two  intersecting  planes,  the  angle  between  these 
perpendiculars  will  he  equal  to  the  dihedral  angle 
hetween  the  planes,  adjacent  to  the  angle  in  which 
the  point  is  situated.  p. 

Hypothesis.  MN,  RS, 
two  planes  (of  which  the 
parts  in  the  diagram  are 
supposed  to  be  rectangu-  ^ 
lar)  intersecting  along  the    "^ 
line  AB;  P,  any  point  in  ^ 
the  obtuse  dihedral  angle 
TBS;  PO,  PQ,  perpen- 
diculars upon  the  planes  from  P. 

Conclusion.     Angle  OPQ  =  dihedral  angle  SBJV. 

Proof.  From  P  drop  the  perpendicular  PC  upon  AB. 
Join  Cd,  CQ.     Then— 

1.  Because  PO  is  perpendicular  to  the  one  plane  and  PQ 
.  to  the  other,  CO  is  the  projection  of  CP  on  the  plane  RS, 

and  CQis  the  projection  of  CP  on  the  plane  ifJV(§§  598, 600). 

Therefore,  J  5  being  perpendicular  to  CP,  the  angles -4  CO 
and  ACQ  are  right  angles  (§  606). 

Therefore  CP,  CO,  and  CQ  are  in  one  plane,  (§  586)  which 
plane  must  also  contain  PO  and  PQ. 

2.  Let  D  be  the  point  in  which  PQ  and  OC  intersect. 

Because,  in  the  triangle  POD,  0  is  a  right  angle, 
— ^1^  nr>n  —  nun  —  «'^T«^i«»»^'ir»+  />-P  /IDD 

=  complement  of  CDQ, 
=  DCQ. 


RELATIONS  OF  PLANES. 


►S 


>'l«l 


299 


Therefore 

Angle  OPQ  =  angle  OCQ, 
4.  Because  CO  and  CQ  are  each  perpendicular  to  A£ 

TZZe    TnToP^^'^hl  r'^  b^etweentlVplanel: 
anereiore       Angle  OPQ  =  dihedral  angle  SBJV,     Q  E  D 

situated  the  point  from  which  they  are  Xppfd  we  J^^fi  '.' 

hem  to  be  supplementary.    Thi/follow  Zm  CcSr^' 

.on^that  the  angle.  SB^  (  =  OOQ)  and  -.^r  Ts^X 

fart^:  TAu,  rt&'  ^^'^'^  '"^  •'"'"*  ^  ^^-'^^ 

perpendicular  upon  the  plane 
MJV  shall  fall  to  the  left  of 
AB,  and  therefore  not  inter- 
sect the  plane  RS.    Then,  if 
we  imagine  ourselves  to  look  M 
directly  along  the  line  AB  so 
as  to  see  both  planes  edgewise, 
the  figure  will   present    this 
appearance. 

whilh^olTteSr  t^  a.-'^^^ralin 
angle.  OPO  and  O^fwilTf  IwoS'tr.le? 
two  angles  will  be  supplementary.     K.  th '     i«    ^ 
dihedral  angle  between  the  planes.  '  °''*™* 

to  0^^."^'°  '"^"'^  "'"^°  *'"'"  *«  ""gl*  ^-^^  is  still  equal 

63'?.  Coroffiary  1."  If  from  any  point  C  of  the  K«,  „f 
xnters^tun  of  two  plane,  two  perpmdiculars:  00  CO  L 
erected,  one  tneach  plane,  andfror^O  and  Q  perpe'ndiXt 


;liii 


300 


BOOK  VIU.    OF  LINES  AND  PLANES, 


Theoeem  XXV. 

639.  ^fa  line  he  perpendicular  to  aplane^  etery 
plane  containing  this  line  is  jlIso  perpendicular  to 
the  plane. 

Hypothesis,    MN,  a  plane ;  PO,  any  line  perpendicular 
to  it,  intersecting  it  in 
0;  ^(7,  any  plane  con- 
taining the  line  PO. 

Conclusion.  The 
plane  AG  i&  perpen- 
dicular to  the  plane 
MN. 

Proof.  From  0 
draw  OD  in  the  plane 
JfiV^  perpendicular  to  ^j5.     Then — 

1.  Because  PO  L  plane  MN,  it  is  perpendicular  to  AB 
and  OD  in  that  plane,  and 

Angle  POD=  right  angle.  (§  584) 

2.  Because  OP  and  OD  are  both  perpendicular  to  AB, 
their  angle  POD  measures  the  dihedral  angle  between  the 
planes  which  intersect  along  AB  (§  623). 

3.  Therefore,  from  (1),  this  dihedral  angle  is  a  right 
angle,  and  the  planes  are  therefore  perpendicular.    Q.E.D. 

Theorem  XXVI. 

630.  If  two  planes  he  perpendicular  to  each  other ^ 
every  line  in  the  one,  perpendicular  to  their  common 
inter  section^  is  perpendicular  to  the  other. 

Hypothesis.  AG, 
MN,  two  plant's  inter- 
secting at  a  right  angle 
along  the  line  AB;  OD, 
a  line  in  the  plane  MN 
perpendicular  to  AB, 

Conclusion.    OD  is 


CI  pCllU.  lU  UXOii        V\J 


cipciiu.iC 

plane  AG, 


.Oil       v\j       v1l\ 


Proof.     In  the  plane  ^(7  draw  OP  1  AB.    Tbon- 


BBLATIOm  OF  PLANm.  g^j 

rtp/i^^""*  ^^  "  perpendicular  to  ^5  (hyp)  ^a  *„ 

Theoeem  XXVII. 
631.  If  two  planes  be  perpendicular  to  each  othpr 

the  line  .4  J9;  OP,  a  line 
perpendicular  to  the  plane 
ABCD. 

Conclusion. 
OP  II  plane  M'JT, 

Proof.  From  anypoint 
R  of  the  plane  MN 
drop  a  perpendicular  i?0 
upon  AB.     Then— 

It  (§607).    Q.ED  '   ^^'  ^^  contains 

fioo    „  "^*  °* '"*^'"««''«<>° -4 A     Hence: 

ott.r.      '''''"''*">''  PorpenatcnUr  to  tU  one  mil  lU  in  the 

»„-    „       Theoeem  xxvni. 


302 


BOOK  nil.    OF  LINES  AND  PLANES. 


Hypothesis.     PQ,  m  two  planes,  each  perpendicxaar  to 
plane  MN;  AB,  their  °  -^ 

line  of  intersection. 

Conclusion. 

AB  L  pla^io  ^^' 
Proof.    1.  Because 
the  plane  PQ  is  per- 
pendicular to  MN,  il 

^rt^^^^^o.  a  perpendicular  to  MN,  it  will 
lie  in  the  P^^ne  P(2  (§  632).  ^^^ 

I-  srorffh^i^i^^^^ 

two'plLes,  or  AB-.  whence^ ^  is  the  perpendicular  to  the 

nXrB-'-  5-%!;Sndicular  to  two  planes 
is  n^Sicular  to  their  line  of  intersection,  and  because  ^1 
plLTiendicular  to  the  same  line  are  parallel  or  com- 
cident  (§  616),  we  conclude: 

634.  All  planes  perpendicular  to  the  same  two  planes  are 
either  parallel  or  coincident. 

Theorem  XXIX. 
636.  Iftwoplanes  are  respectively  perpendicular 
to  two  intersecting  Ivnes,  their  Une  of  inter sect^on  ^s 
perpendiealar  to  the  plane  of  the  lines. 

Hypothesis.    OH,  Oh  two  lines  intersecting  at  0; 
Plane  UN  1  line  OJI\ 
Plane  KL  1  line  0/; 
TJV,  the  line  of  inter- 
section of  these  planes. 
Conclusion. 

VY  L  plane  EOI. 
Proof.  1.  Because  the 
plane  MN  is  perpendicular 
to  OH,  it  is  perpendicular 
to  every  plane  passing 
through  OH.    That  is, 

Plane  MN 


RELATIONS  OF  PLANES. 


303 


2.  In  the  same  way^ 

Plane  KL  ±  plane  HOI, 

3.  Because  ea<5h  of  the  planes  is  perpendicular  to  HOL 
and  UViB  their  line  of  intersection, 

?7r  X  plane  ^0/ (§  633).     Q.E.D. 

Belatlons  of  Three  or  more  Planes. 
636.  Remark.    When  three  planes,  which  we  may  call 
^,  r,  and  Z,  mutuaUy  intersect,  there  will  be  three  lines  of 
intersection: 

One  line  formed  by  the  planes  X  and  F; 
One  line  formed  by  Fand  Z; 
One  line  formed  by  Z  and  X. 


Theoeem  XXX. 

637.  The  three  lines  of  intersection   of  three 
planes  are  either  parallel  or  meet  in  a  point. 

also'^Ilf'    ^*  ""^  ^^  *^^  ^^""^^  ^^^""^^  '^^  ^'  ^""^  ^*  ^^'^  ^® 
a,  the  line  of  intersection  of  Xand  Y' 
h,  the  line  of  intersection  of  Fand  zl 
c,  the  line  of  intersection  of  Z  and  X 

Then— 

1.  Because  the  lines  a  and  5  both  lie  in  the  plane  T 
they  are  either  parallel  or  intersect  each  other.  The  same 
may  be  shown  for  i  and  c,  and  for  c  and  a. 

3.  Suppose  a  and  b  to  intersect.  Because  a  lies  in  both 
the  planes  Xand  F,  and  h  lies  in  both  Fand  Z,  the  point 
where  they  intersect  must  lie  in  all  three  planes  X,  f;  and 
Z  Therefore  it  must  lie  on  both  the  planes  Xand  Z,  and 
therefore  on  their  line  of  intersection  c.  The  three  lines  a, 
b,  and  c  will  then  all  meet  at  this  point. 

3.  If  a  and  b  are  parallel,  c  cannot  meet  either  of  them, 
because,  by  (2),  where  it  meets  the  one  it  must  meet  the 
otuer  aiBo.    xnerefore,  in  this  case,  none  of  the  lines  will 

V^ZI^T^  ^x?  ""^  *^®  ''^^®^''  *^^  ^6ca«se  each  pair  lies  in  the 
same  plane  they  must  be  all  parallel. 


i   " 


Ih!* 


304  BOOKVIIL    OF  LINE8  AND  PLANES, 

«Qft    Cnrollar'U  1.    //  the  three  lines  of  intersection  of 
J^ne'^:^^^^^^  ^/-  ^^ree planes  all  pass  through 

thatpmnt,  ^^^  ^^^^^  ^^^  ^^^^^^^^  ^^  ^^^^  ^^^^ 

of  mm  is  also  pa/allel  to  the  intersection  of  any  two  planes, 
one  of  which  passes  through  each  of  the  lines. 

Theorem  XXXI.  v 

640.  If  the  Ivnes  of  intersection  of  three  planes 
are  parallel,  any  fourth  plane  Verpendi^^^^^ 
of  the  three  planes  is  also  perpendicular  to  the  mra 
^  HypothesL    The  parallel  lines  a,  h  o  are  the  lines  of 
intersection  of  three  planes,  « 

which  wo  ckW  the  planes  a&, 
Ic,  and  cff;  MN,  a  plane 
perpendicular  to  the  two 
planes  al  and  Ic. 

Conclusion.    ifiV  is  also 
perpendicular  to  the  plane 

ac. 

Proof     1.    Because  the  . 

plane  uk  is  perpendicular  to  both  the  planes  a&  and  &.,  it  is 

T^prnendicular  to  their  line  of  intersection  I  (§  633). 

^  TBecau^  MNi.  perpendicular  to  5,  it  is  perpendicular 

to  the  lines  a  and  c,  parallel  to  5  (§  591). 

3.  Therefore  it  is  perpendicular  to  the  plane  ac,  wnicn 

passes  through  those  lines  (§  629).     Q.E.D. 
Scholium.    The  most 

remarkable   position   of 
three  planes  is  that  in 
which  each  plane  is  per- 
pendicular to  the  other 
two.    By  the  preceding 
theorems   each   line    of 
:^4^,v^n/>o+irkTi  will  he  per- 
pendicular  to  the  other 
two   lines    of    intersec- 
tion and  to  the  third  plane. 


i!  . ^,„,,„... 


POLYHEDRAL  ANQLB& 


806 


n  of 
mgh 

each 
ineSf 


anes 
>  two 
ird. 
les  of 


c,  it  is 
dicular 
,  which 


CHAPTER    III. 

OF  POLYHEDRAL  ANGLES. 


A  polyhedral  angle. 
O,  the  vertex;  OA,  OB.  00, 
etc.,  the  edges ;  AOB,  BOO, 
etc.,  the  faces. 


641.  Def,  When  three  or  more  planes  pass 
through  the  same  point,  they  are  said  to  form  a  poly- 
hedral  angle  at  that  point. 

A  polyhedral  angle  is  also  called  a 
solid  angle. 

Each  plane  which  forms  a  poly- 
hedral angle  is  supposed  to  be  cut  off 
along  its  lines  of  intersection  with  the 
planes  adjoining  it  on  each  side.  "|    /  \     "'>C 

642.  Def,  Edges  of  a  poly- 
hedral angle  are  the  straight  Jines 
along  which  the  planes  intersect. 

643.  Def.  Faces  of  a  poly- 
hedral angle  are  the  planes  which  form  it. 

644.  Def.  The  vertex  of  a  polyhedral  angle  is 
the  point  where  the  faces  and  edges  all  meet. 

645.  The  edges  of  a  polyhedral  angle  may  be  produced 
indefinitely.  But  to  make  the  study  of  the  angle  easy,  the 
faces  and  edges  may  be  supposed  cut  off  by  a  plane.  The 
intersection  of  the  faces  with  this  plane  will  then  form  a  poly- 
gon, as  ^^(72)^. 

This  polygon  is  the  base  of  the  polyhedral  angle. 

646.  Each  pair  of  faces  which  meet  an  edge  form  a 

dihedral  angle  along  that  edge.    There  are  as  many  edges  as 

faces,  and  therefore  as  many  dihedral  angles  as  faces. 

Hence  two  classes  of  angles  enter  into  any  polyhedral 
angle,  namely: 

I.  The  plane  angles  AOB,  BOO,  COD,  etc.,  called  also 
face  angles,  which  the  edges  form  with  each  other.  The 
planes  on  which  these  angles  are  measured  are  the  faces. 


806 


BOOK  VUL    OP  LINSa  AND  PLANES. 


i 


Example.  The  plane  of  the  angle  A  OB  is  the  face  bounded 

by  OA  and  OB.  „  a    , 

II.  The  dihedral  angles  between  the  faces,  called  awo 
edge  angUs.  By  §  623  each  of  these  angles  is  measured  by 
the  plane  angle  between  two  lines,  one  in  each  face  perpen- 
dicular  to  the  edge  of  the  dihedral  angle. 

If  the  cutting  plane  ABODE  vfQXQ  perpendicular  to  one  of 
the  edges,  say  OB,  then  the  dihedral  angle  along  OB  would 
be  measured  by  the  plane  angle  ABG. 

But  this  plane  cannot  be  perpendicular  to  more  than  one 
edge,  so  that  to  measure  the  dihedral  angles  in  this  way  we 
must  have  as  many  cutting  planes  as  edges. 

647.  Def.  Two  polyhedral  angles  are  identioally 
equal  when  they  can  be  so  applied  to  each  other  that 
al  the  faces  and  edges  of  the  one  shaU  coincide  with 
the  corresponding  faces  and  edges  of  the  other. 

In  order  that  such  coincidence  may  be  possible,  the  face 
and  dihedral  angles  of  the  one  must  all  be  equal  to  the  face 
and  dihedral  angles  of  the  other,  taken  in  the  same  order,  each 
to  each. 

Positive  and  Negative  Rotations. 

648.  When  a  person  looking  down  upon  a  point  0  sees 
a.  motion  around  that  point  in  a  direction 
the  opposite  of  that  of  the  hands  of  a 
watch,  the  motion  is  said  to  be  positive 
relative  to  his  standpoint. 

If  the  motion  is  in  the  other  direction,  I 
it  is  said  to  be  negative. 

A  motion  which  is  positive  when  seen 
from  one  side  will  appear  negative  when 
the  observer  views  it  from  the  other  side      Positive  rotation, 
of  the  plane,  or  when  the  figure  is  turned  over  so  as  to  be 
seen  from  the  other  side. 

For  illustration,  imagine  one's  self  seeing  the  hands  of  a  watch  by 
looking  through  it  from  behind. 

To  avoid  ambiguity  one  side  of  the  plane  of  motion  may 
be  taken  as  positive  and  the  other  side  as  negative.    Then  a 


POLYHEDRAL  AKQLSa. 


d07 


positive  rotation  is  that  which  appears  positive  when  seen 
from  the  positive  side,  or  negative  as  seen  from  the  negative 
side. 

Astronomical  illustration.  If  one  could  look  down  upon 
the  earth  from  above  the  north  pole,  the  earth  would  appear 
rotating  in  the  positive  direction.  If  he  should  look  down 
upon  it  from  above  the  south  pole,  it  would  appear  rotating 
in  the  negative  direction. 

Remark.  The  habit  of  regarding  the  motion  opposite  that  of  the 
hands  of  a  watch  as  positive  arose  from  the  direction  of  the  north  pole 
being  taken  as  positive,  because  astronomy  was  developed  among  the 
people  of  the  northern  hemisphere;  these  people  regarded  as  positive 
the  direction  in  which  tlie  earth  would  appear  to  rotate  when  seen  from 
the  north. 

649.   As  there  are  positive  and  negative  rotations,  so 
letters,   angles,   and  lines 
may  succeed  each  other  in 
a  positive  or  negative  di-^ 
rection. 


650.  NoTATioia".  A 
polyhedral  angle  is  desig- 
nated by  a  letter  at  its  ver-  letters  succeed  each 

tex  followed  by  a  hyphen  S.""  "^^  "^"'^ 
and  the  letters  at  the  vertices  of  its  base. 


Letters  succeed  each 
other  in  the  negative 
ordor. 


661.  Def.  Symmetrical  polyhedral  angles  are 
those  which  have 
their  plane  and  di-  "^ 
hedral  angles  equal, 
each  to  each  but 
arranged  in  reverse 
order,  the  one  posi- 
tive and  the  other 
negative  when  seen 
from  the  vertex. 


B  BT- 


Example.  The  tri-  ^' 
hedral  angles  0-ABG  and  O'-A'B'C*  are  symmetrical  when 


\ 


't 


I 


308  BOOK  nil    OF  LINES  AND  PLANES. 

Angle  ^Oi?  =  angle  A'0'B\ 
"  BOC=z  "  B'0'C\ 
"      00  A  =     "     CO' A', 

Dihedral  angle  along  edge  OA  =  dihedral  angle  along  edge  O'A ', 
"        "  ,     **       **    0B=        "        "        "      *<   O'B' 
"        "        '<       "    00=        "        "        "      "   O'c' 

The  two  symmetrical  polyhedral  angles  may  be  so  cut  that 
the  base  ABO  shall  be 
identically  equal  to  the  ^^ 

base  ^'i?'C".  But,  in 
order  to  bring  these 
bases  into  coincidence, 
one  of  the  figures  must 
be  turned  over  and  the 
bases  brought  together 
with  the  vertices  in  opposite  directions,  as  in  the  figure. 

Hence  two  symmetrical  polyhedral  angles  cannot  in  gen- 
eral bo  brought  into  coincidence. 

653.  Be/.  Opposite  polyhedral  an- 
gles are  those  each  of  which  is  formed 
by  the  continuation  of  the  edges  and 
faces  of  the  other  beyond  the  common 
vertex. 

Example.  If  the  lines  AO,  BO,  00,  and 
DO  are  produced  through  0  to  the  respective 
points  A',  B',  C,  D',  then  the  polyhedral ^< 
angle   0-A'B'C'D'  is   the  opposite  of   the 
angle  O-ABOD, 


Theorem  XXXII. 

663.  Opposite  polyhedral  angles  are   symmet- 
rical. 

Hypothesis.  0-^ 5  Ci),  any  polyhedral  angle:  O-A'B'O'D' 


POLYHEDHAL  ANGLES. 


809 


Conclusion. 

Angle  0-A'B'C'D'  is  symmetrical  to  O-ABCD, 

Proof.  1.  Because  ^0^1' 
and  BOB'  are  in  the  same 
straight  line, 

Pace  angle  A' OB'  =  opp. 
angle  A  OB. 
In  the  same  way  it  may  bo 
shown  that  every  other  face 
angle  of  the  one  is  equal  to 
the  opposite  face  angle  of  the 
other.  ^ 

2.  Because  the  linos  ^0^' and  BOB' ^kbb  through  the 
same  point,  they  are  in  the  same  plane.  Therefore  the  face 
A  OB'  IS  in  the  same  plane  with  the  face  AOB.  In  the 
same  way,  every  other  pair  of  corresponding  faces  are  in  the 
same  plane. 

3.  Because  the  dihedral  angle  between  two  planes  is  every- 
where the  same  (§  619),  each  of  the  edge  angles  OA',  OB', 
etc.,  IS  equal  to  the  corresponding  one  of  the  edge  angles 
OA,  OB,  etc.,  of  the  other  angle. 

4.  If  one  should  look  down  upon  the  figure  from  above, 
the  letters  ABCD  and  A'B'O'D'  would  each  succeed  each 
other  in  the  positive  order.  Hence  if  the  opposite  angle  is 
turned  over  into  the  position  0-A"B"C"D",  the  order  of  the 
letters  wiU  appear  negative,  and  therefore  the  opposite  of 
those  in  the  original  polyhedral  angle. 

5.  Therefore  the  two  polyhedral 
angles,  being  equal  in  all  their  face  and 
edge  angles,  but  having  them  arranged 
in  reverse  order,  are  symmetrical. 

Q.E.D. 
654.  Def.  A  trihedral  angle  is 
a  polyhedral  angle  which  has  three 
edges,  and  therefore  three  faces. 

Remark.     In  a  trihedral  angle  each ^^^v^^,^. 

tsxie  has  an  opposite  edge,  and  each  ed^e  K^^^gj!  °PP!t*~^^- 
an  opposite  face.  ^  ^'ace  obc  is  opp.  ^gi  oZ 


A  trihedral  angle. 


310 


BOOK  VIII.    OF  LIMES  AND  PLANES, 


^--C 


Theorem  XXXIII. 

655.  jj/  two  face  angles  of  a  trihedral  angle  are 
equal,  the  edge  angles  opposite  them  are  also  equal. 

Hypothesis.     0-ABO,  a  trihedral  angle  in  which  face 
angle  BOA  =1  face  angle 
BOG. 

Conclusion.  Edge  angle 
OA  —  edge  angle  OC. 

Proof.  From  any  point 
P  of  OB  drop  the  perpen- 
diculars PM  _L  OAi 

PN  L  00; 

PD  1  plane  ^Oa 
Join  DM,  DN.    Then— 

1.  Because  PM  intersects  the  plane  A  OC  in  Jf,  and  PD 
is  perpendicular  to  this  plane,  MD  is  the  projection  of  PM 
upon  this  plane  (§  600).    Therefore,  because  OA  1  PM, 

OA  1  MD.  (§606) 

2.  Because  MD  and  MP  are  pependicular  to  OA  in  the 
planes  forming  the  dihedral  angle  OA, 

Dihedral  angle  OA  =  plane  angle  PMD.      (§  623) 

3.  In  the  same  way,  we  show 

Dihedral  angle  OC  =  plane  angle  PJVD. 

4.  In  the  right-angled  triangles  POM  and,  PON, 

The  side  OP  is  common, 
Angle  PON  =  angle  POM  (hyp.); 
therefore  these  triangles  are  identically  equal,  and 

PN=  PM. 
6.  In  the  right-angled  triangles  PDN&nd  PDM, 
Side  PD  is  common, 
PN  =  PM;  (4) 

therefore  these  trianglen  are  identically  equal,  and 
Angle  PND  =  angle  PMD. 
6..  Comparing  this  result  with  (2)  and  (3), 
Dihedral  angle  OA  =  dihedral  angle  OC.     Q.E.D. 


POLTHBDBAL  ANGLES. 


311 


(4) 


Theorem  XXXIV. 

656.  In  a  trihedral  angle  the  greater  face  angle 
and  the  greater  edge  angle  are  opposite  each  other. 

Hypothesis.     0-ABC,  a  trihedral  angle  in  which  foce 
angle  BOA  >  face  angle 
BOO. 

Conclusion.  Edge  angle 
OC  >  edge  angle  OA. 

Proof.  Make  the  same 
constructions  as  in  the  last 
theorem.     Then — 

1.  In  the  same  way  as 
in  the  last  theorem  fol- 
lows: 

Edge  angle  OA  =  plane  angle  PMD. 
Edge  angle  00  =  plane  angle  PND. 

2.  In  the  right-angled  triangles  POJfandPOiV  the  line 
OP  IS  the  common  hypotheuuse;  and  because  angle  POM  is 
greater  than  angle  PON, 

Line  PM  >  PN. 

3.  In  the  right-angled  triangles  PDNm^  PBM,  because 
the  side  PB  is  common  and  the  hypothenuse  PM  greater 
than  the  hypothenuse  PJV, 

Angle  PJ^B  >  angle  PMB. 

4.  Comparing  this  result  with  (1), 

Edge  angle  00  >  edge  angle  OA.    Q.E.D. 
Corollary  1.     From  these  two  theorems  it  follows,  as  in 
Bk.  II.,  §115: 

657.  If  one  edge  angle  is  greater  than  another,  the  face 
angle  opposite  it  is  greater  than  that  opposite  the  other. 

For  the  face  angles  could  not  be  equal  without  violating 
Theorem  XXXIIL;  nor  could  that  opposite  the  lesser  edge 
angle  be  the  greater  without  violating  Theorem  XXXIV. 

658.  Cor.  2.  If  the  edge  angles  be  arranged  in  the 
order  of  magnitude,  the  face  angles  opposite  them  will  be  in 
the  same  order  of  magnitude,  so  that  the  smallest,  mean,  and 
greatest  angle  of  the  one  class  will  be  opposite  the  smallest, 
mean,  and  greatest  angle  of  the  other,  respectively. 


im 


i 


312 


BOOK  VIU.    OF  LINES  AND  PLANBB. 


Theorem  XXXV. 

659.  In  a  trihedral  angle  in  which  each  of  the 
face  angles  is  less  than  a  straight  angle,  the  sum  of 
any  two  face  angles  is  greater  than  the  third. 

Hypothesis.  0-ABG,  a  trihedral  angle  in  which  AOG  la 
the  greatest  face  angle. 

Conclusion.  The  sum  of  the  face 
angles  A  OB  and  BOO  is  greater  than 
AOC. 

Proof.  Through  0  draw  in  the 
plane  A  00  a,  line  OD,  making  angle 
AOD  =  angle  A  OB.  Let  the  base 
ABC  he  so  cut  off  that  we  shall  have 
OB  =  OD^    Then— 

1.  Because  the  triangles  OAB  and  OAD  have 

Side  OA  common, 
Side  OD  =  OB,     )  ^^,  „i.^,«+:«„ 
Angle  ^  05  =  ^OAr"'^'*^"*'^^' 
they  are  identically  equal,  and 

AB  =  AD. 

2.  Because  ^5C  is  a  plane  triangle. 

Sides  AB  +  BC>  third  side  A G. 

3.  Taking  away  from  this  inequality  the  equal  lengths  AB 
and  AD, 

BC>DC. 

4.  Because,  in  the  two  triangles  OCB  and  OCD, 

Side  OCia  common, 

OD  =  OB  (const.), 
and 

CB  >  OD, 
we  haye 

Angle  BOC  >  angle  OOD. 

5.  Adding  the  equal  angif^s  A  OB  and  A  OD, 


(§115) 


Angle  AOB  +  angle £00 >  angle  A 00.    Q.E.D. 


POLTHEDhAL  ANGLB8. 


313 


Theorem  XXXVI. 

660.  Two  trihedral  angles  are  either  equal  or 
symmetrical  when  the  three  face  angles  of  the  one 
are  respecUvely  equal  to  the  face  angles  of  the  other. 

Hypothesis.  0-ABC,0''A'B'G',  0"-A"B"C'',  three  tri- 
hedral angles  in  which 

Face  angle  AOB  =  angle  A'O'B'  =  angle  A"0"B"- 
"  "  BOG=  "  B'Q'Q'-  «  B''0"G"'' 
"        "     COA  =      «    G'O'A'  =     "      C"0"^"- 

orders  of  angles  ABG  and  A'B'G'  positive  when  viewed 

from  the  vertex,  and  of  A"B"G''  negative. 


Jf  elusion  The  edge  angles  of  the  trihedral  angles  are 
also  equal,  and  the  trihedral  angles  O-ABC  and  0'  A'B'C' 
are  equal,  and  0".A"B-G-  is  symmetrical  with  them. 

fakP  r-^*     ^^*^.'  f ''  ^^^  ^'^''  ^""^  ^"^"^  respectively, 
take  the  equal  distances    OP,   O'P'    and   r)"P"    o«^ 

1.  In  the  triangles  OPQ,  0'P'Q\  0"P''Q" 

A     1    „^^  =  ^'^' =  ^"-^"  (const.).' 
Angle  P(9^  =  P'0'^'  =  P"0"0"(hvp) 
OPQ  =  O'P'^'  =  0"i>"r  (all  being  rigKngles). 

Therefore  these  triangles  are  identicaUy  equal,  so  that 

OQ  =  O'Q'  =  0''Q", 
PQ=P'Q'=.P''Q'\ 


siiy 


314 


BOOK  Vin.    OF  LINES  AND  PLANES. 


2.  In  the  same  way,  OPR,  O'P'R',  and  0"P"R"  being 
right  angles  by  construction, 

OR  =  O'R'  =  0"R'\ 
PR  =  P'R'  =  P"R'\  V 

Imagine  QR^  Q'R't  and  Q"R"  to  be  joined,  then — 

3.  In  the  triangles  OQR,  0'Q'R\  0"Q"R",  etc.. 

Face  angle  QOR=  Q'O'P'  =  Q"0"R"  (hyp.), 
and  the  sides  which  include  these  angles  are  equal  by  (1)  and 
(2).     Therefore  these  triangles  are  identically  equal,  and 

QR  =  Q'R'  =  q"R' 


vt 


4.  Comparing  with  (1)  and  (2),  the  three  triangles  PQR, 
P'Q'R'j  P'*Q*'R*'  have  their  corresponding  sides  equal,  and 
are  therefore  identically  equal  to  each  other.     Hence 

Angle  qPR  =  angle  Q'P'R'  =  angle  Q"P**R'', 

5.  Because  QP,  QR,  etc.,  are  each  perpendicular  to  their 
edges,  and  lie  in  their  respective  faces,  their  angles  measure 
the  dihedral  angle  between  those  faces.     Therefore 

•  Edge  angle  OA  =  edge  angle  O'A^  =  edge  angle  O'M". 

6.  In  the  same  way  may  be  shown 

Edge  angle  OB  =  edge  angle  O'B^  =  edge  angle  0"J5". 
Edge  angle  00  =  edge  angle  O'O'  =  edge  angle  0"(7". 
Therefore  the  three  trihedral  angles  have  their  edge  angles 
all  respectively  equal. 

7.  Because  in  the  first  two  trihedral  angles  the  arrange- 
ment of  the  equal  angles  is  the  same,  while  in  the  third  this 
arrangement  is  reversed,  the  first  two  angles  are  equal,  and 
the  third  symmetrical  with  them.    Q.E.D. 


1 

0 


POLYHEDRAL  ANGLES. 


being 


316 


Theorem  XXXVII. 

fnl^^'  1^  ?  ^^'^'^^^  P^^^y^edral  angle  the  sum  of  the 
/ace  angles  ts  less  than  a  perigon  (360°). 

thef^lf  tfi.    O.^^C/>^,  a  polyhedral  angle  of  which  aU 
the  angles  of  the  base  are  convex. 

Conclusion.       Angles    AOB  -\-  BOG 
+  etc.  +  EOA  <  360°. 

Proof.     Let  n  be  the  number  of  faces 
A^nr.  f"^"^  polyhedral  angle.     The  base 
ABLDEmW  then  be  a  polygon. of  n  sides, 
l^t  us  also  put  2,  the  sum  of  the  face  A 
angles  A  OB,  BOO,  etc.     Then— 

1.  Because  ABODE  is  a  convex  poly- 
gon of  n  sides,  ^ 

Angle  ABO  +  angle  BCD  +  etc.  =  („  -  3)  straight  angles. 

anie«  ^fT""  t""*  ^T  *°™  "  *™°SH  the  sum  of  alMhe 
angles  of  these  triangles  is  n  straight  angles.     That  is, 
2  +  angles  {OAB  +  05^  +  OBG->t  00B+  etc.) 

=  «  straight  angles. 

ande  a^Tof  wr  tTf'^^'  f  ^^'  ^^^form  a  trihedral 
angle  at  B,  of  which  the  face  angles  are  OBA    OBO  ABC 

Angle  OBA  +  OBO  >  J^a 
Angle  OC^  +  OCD  >  ^CZ).  (§  659) 

etc.        etc.        etc. 
4.  Taking  the  sum  of  these  inequalities,  we  find 
Sum  of  the  %n  base  angles  of  triangles  OAB,  OBO  etc 
^eater  than  the  sum  of  the  angles  of  the  polygon  ABODE- 
that  18,  greater  than  n-^  straight  angles. 

resi"(.7an'dVi:'  *^  ^""  "'  '^'^  "^  -^les,  the 

-^  +  -^  =  w  straight  angles. 
^  >  (/i  —  2)  straight  angles. 
The  difference  of  these  shows  that 


or 


2  <1  2  Strftiorlif,  anncl 


Sxv,o, 


^<360°    Q.E.D. 


I'* 


' .  1 4  I 


Jf 


n 


;Fi«i  Fiffi 


BOOK  IX. 
OF  POLYHEDRONS. 

CHAPTER   I.  .      , 

OF  PRISMS  AND  PYRAMIDS. 


:ll'IIlll{ 


662.  Definition.  A  solid  is  that  which  has  length, 
breadth,  and  thickness. 

A  solid  is  bounded  by  a  surface. 

Eemark.  The  form,  magnitude,  and  position  of  a  solid 
are  completely  determined  by  the  form,  magnitude,  and 
position  of  its  bounding  surface.  Hence  we  may  consider 
the  surface  as  defining  the  solid.  « 

663.  Def.  A  polyhedron  is 

a  solid  bounded  by  planes. 

664.  Def.    The  faces  of  a 

polyhedron   are   its    bounding  A< 
planes. 

665.  Def.    The  edges  of  a 

polyhedron    are    the   lines   in 

which  its  faces  meet.  ^ 

A  polyhedron. 

666.  Def.    The  vertices  of   The  planes  hab,  hbc,  akb, 

t    ■,      t  ,-,  .    .      .     6tc.,  are  the   faces.     The  lines 

a  polyhedron  are  the  points  m^^.^^,  etc.,  bounding  the  faces, 

i~.   T_  . ,         ,  ,        -^  are  the  edges.    The  points  A,  B, 

which  its  edges  meet.  C,  D,  H,  K&re  the  vertices. 

667.  Def.  Diagonals  of  a  polyhedron  are  straight 
lines  joining  any  two  vertices  not  in  the  same  face. 

668.  Def  A  plane  section  of  a  polyhedron  is 
the  polygon  in  which  its  faces  cut  a  plane  passing 
through  it 


PRISMS. 


317 


669.  I)ef.  Two  polygons  are  said  to  be  paraUel 
to  each  other  when  each  side  of  the  one  is  paraUel  to 
a  corresponding  side  of  the  other. 

Prisms. 

670.  Def.  A  prism  is  a  polyhedron  of  which  the 
end  faces  are  equal  and  parallel  polygons,  and  the 
side  faces  parallelograms.  ^o      > 

em.  De/:    The  bases  of  a  prism  are  its  end  faces. 

673.  Def.  The  lateral  faces  are  aU  except  its 
pases. 

673.  Be/,   The  lateral  edges 

of  a  prism  are  the  intersections  of 
its  lateral  faces. 

674.  Def.  A  right  section  of 

a  prism  is  a  section  by  a  plane 
perpendicular  to  its  lateral  edges. 

675.  Def.  A  prism  is  said  to 
be  triangular.  Quadrangular  h«»    .  „f°  hexagonal  prism. 

»5»ufu,«iuaurailgUiar,neZ-   ABCDEF  and  A'B'C'iyE'S^ 

agonal,  etc. ,  according  as  its  bases        *^  ***«  '^^^  ^^'^^ 
are  triangles,  quadrilaterals,  hexagons,  etc. 

;,.  ^J^^'P^f'    The  altitude  of  a  prism  is  the  perpen- 
dicular distance  between  its  faces. 

1  ^^Vl  -^^^'    ^  ^^"^  P^^"^  i»  one  in  which  the 
lateral  faces  are  perpendicular  to  its  bases. 

1  ^^"^h  ^^^'    ^^  oblique  prism  is  one  in  which  the 
lateral  faces  are  not  perpendicular  to  the  bases. 

679.  Def.  A  regular  prism  is  a  right  prism  whose 
bases  are  regular  polygons. 


;i.i-;;ii 


M 


mk. 


jk.«Bl 


318 


BOOK  IX.     OF  POLYHEDJiONS. 


! 


Ill 


TlIEORSM  I. 

680.  The  lateral  edges  of  a  prism  are  equal  an4 
parallel,  and  make  equal  angles  with  the  bases. 

Hypothesis.    ABC,  A'B'C,  two  edges  of  the  bases  of  a 
prism;   ABA'B',  BCB'C,  the  lateral 
faces  joining  those  edges. 

Conclusion.  AA',  BB',  CC  are 
equal  and  parallel  and  make  equal 
angles  with  the  bases. 

Proof.  1.  Because  ABA'B'  is  a 
parallelogram  (§  670), 

Line  BB'  -  and  ||  AA', 

2.  Because  BB'CC  is  a  parallelo- 
gram,       ! 

Line  BB'  =  and  ||  CC\ 

3.  Because  AA'  and  CC  are  equal  and  parallel  to  the 
same  line,  they  are  equal  and  parallel  to  each  other  (§  592). 

4.  In  the  same  way  it  may  be  shown  that  all  the  other 
lateral  edges  are  equal  and  parallel.    Q.E.D. 

5.  Because  the  lateral  edges  are  parallel  lines,  they  make 
equal  angles  with  either  base  (§  610).     Q.E.D. 

6.  Because  the  bases  are  parallel  planes,  the  edges  make 
equal  angles  with  the  two  bases  (§  618).     Q.E.D. 

Theoeem  II. 

681.  The  sections  of  the  lateral  faces  of  a  prism 
by  parallel  planes  are  equal  and  parallel  polygons. 

Hypothesis.     ABCD-A'B'C'D\   a  prism;    EFOH  and 
E'F'G'H,  sections  of  the  lateral  faces  by  parallel  planes. 
Conclusion.    EFGH  =  and  ||  WFG'W. 
Proof.   1.  Because  A  A'  and  BB'  are  parallel  lines  inter- 
secting parallel  planes, 

EE'  =  and  \\_HF',    _  _  _  (§  615) 

Therefore  the  four  lines  EF,  FF%  F'E%  and  E'E  form  a 
parallelogram  and 

Line  ^ii^=  and  \\E'F'. 


PIiI8M% 


tal  an(i 

es. 

a^es  of  a 


^c 

b1  to  the 
§  592). 
the  other 

bey  make 

ges  make 


a  prism 
ygons. 

'OH  and 

lanes. 

nes  inter- 

(§615) 
'E  form  a 


3.  In  the  ame  way  we  find 


819 


Line  FG  =  and  ||  ro'. 

Line  OH—  and  ||  0'H\ 

etc.  etc. 

A     ^1%^}^  '^^®'  ^^  *^®  respective  angles  are  paraUel 
Angle  BFO  =  angle  E'F'O',  parauei. 

Angle  FOH  =  angle  F'0'h\ 

,  (§  608)       ^^ 

etc.  etc.  ^ 

Therefore  the  polygons  EFOH 
and  ^'^'(?'^',   having    their  sides    t\ 
and  angles,  taken  in  order,  all  equal 
are  identically  equal.     Q.E.D.  ' 

683.  Corollary,   Any  section  of  */ 
a  prism  hy  a  plane  parallel  to  the  d 
bases   is    identically    equal    to   the 
oases. 

Theorem  III 

1.7  T^T.'  ff'^^/'f^'^  of  a  prism  hy  a  plane  paral- 
lel to  the  lateral  edges  is  a  parallelogram. 

Hypothesis.     ABCA'B'G^   any    prism:   PORS   a  sec 
*Z,'^  *^^'  P"sn^  by  a  plane  parallel  to      ^  ^     '        '" 
AA'y  BB'.  *X zpiTi 

Conclusion.     PQRS  is  a  parallelo- 
gram. 

Proof.   1.  Because  the  line  AA'  is 

parallel  to  the  intersecting  plane  PQRS, 

it  cannot  meet  either  of  the  lines  PR  or 

Q8  which  lie  m  that  plane.     Therefore 

.  the  lines  A  A',  PR,  and  QS  are  parallel.  ^ 

(§  637) 
3.  Because  the  opposite  sides  AP         jl 

^''^^f^^:.  ^^'  ^""^  ^^  ^^  *be  quadrilateral  APA'R  are 
parallel,  the  quadrilateral  is  a  parallelogram,  and 

„    ,    ,,  ^^  =  and  II  AA\ 

o.  m  the  same  way  we  may  prove 

..    rpu      .         ^^""^  QS  =  B.n^\\  A  A'. 

PO^^t  ii^'^"^"^  "  ^^'   ^^^^    *be    quadrilateral 

i^^//^6  IS  a  parallel o^rnni.     Q.E.D. 


"f 


320 


BOOK  IX.     OF  POLYHEDRONS. 


Parallelepipeds. 

684.  Def.  A  parallelopiped  is  a  solid  contained 
by  three  pairs  of  parallel  planes. 

A  parallelopiped  is  therefore  a  prism  of  which  the 

lateral  faces   are   two  pairs  of 
parallel  planes. 

685.  Def.  A  rectangular 
parallelopiped  is  one  whose 
faces  intersect  at  right  angles. 

686.  D^.  A  cube  is  a  par- 
allelopiped whose  faces  are  all 

squares.  ^  paraUeloplped. 

Theorem  IV. 

687.^  The  opposite  faces  of  a  parallelopiped  are 
identically  equal  parallelograms. 

Hypothesis.     ABCD-BFGH,  any  parallelopiped. 

Conclusion.     The  opposite  faces  ABCD  and  BFOH  are 
identically     equal      parallelo- 
grams. 

Proof.  1.  Because  5(7  and 
AD  are  the  lines  in  which  the 
parallel  planes  BCFG  and 
ADEH  intersect  the  third 
plane  ADBO, 

BG  II  AD.      (§  614) 

2.  In  the  same  way  it  may 
be  shown  that  the  lines  AB  and  DC  are  parallel.      Therefore 

ABGDiBVi,  parallelogram.     Q.E.D. 

3.  It  may  be  shown  in  the  same  way  that  all  the  other 
faces  are  parallelograms.  Therefore,  by  comparing  opposite 
sides  of  successive  parallelograms, 

AB  =  and  ||  EF, 
and  5C  =  and  II  i^(?. 

4.  Because  the  sides  BA  and  BG  ot  the  angle  ABG  bxq 
parallel  to  the  sides  FE  and  FG  of  the  angle  EFGy 

Angle  ABG  =  angle  EFG.  (§  608) 


PAUALLEL0P1PED8. 


821 


Qtalned 
ich.  the 


led. 


oed  are 


therefore 

.he  other 
opposite 


iBG  are 
(§  608) 


5.  Therefore  the  parallelograms  A  BCD  and  EFQII 
having  their  respective  sides  and  one  angle  equal,  are  identi- 
cally equal.  In  the  same  way  it  may  be  shown  that  every 
other  pair  of  opposite  faces  are  equal.     Q.E.D. 

Corollary  1.  The  edges  of  a  parallelopiped  are  twelve  in 
number y  and  may  be  divided  into  three  eets,  each  set  compris- 
ing/our equal  and  parallel  lines. 

Cor.  2.   The  vertices  of  a  parallelopiped  are  eight  in  number. 

Cor.  3.  The  diagonals  of  a  parallelopiped  are  four  in  num- 
ber, and  may  be  drawn  from  any  angle  of  each  face  to  the 
opposite  angle  of  the  opposite  face. 

Theorem  V. 

688.  The  four  diagonals  of  a  parallelopiped  all 
intersect  in  a  point  which  bisects  them  all. 

Hypothesis.    ABCD-EFGH,  any  parallelopiped. 

Conclusion.    The  four  diagonals  ^^ 

AGy  BH,  CE,  and  i>i?^  all  inter- ^[ 
sect  in  a  point  (9,  and  are  bisected 
by  this  point. 

Proof.  Through  the  opposite 
parallel  edges  AB  and  HG  pass  a 
plane.     Join  AH,  BG.     Then — 

1.  Because  the   sides  AB  and  -^ 
HG  of  the  quadrilateral  ABHG  are 
equal  and  parallel,  ABHG  is  a  par- 
allelogram.    Therefore  AG  and  BH,  the  diagonals  of  this 
parallelogram,  intersect  and  bisect  each  other. 

Let  0  be  the  point  of  intersection. 

2.  In  the  same  way  it  may  be  shown  that  BH  and  CE 
bisect  each  other.  Therefore  CE  passes  through  the  point  of 
bisection  0,  and  0  bisects  CE. 

3.  In  the  same  way  it  may  be  shown  that  DF  passes 
through  0  and  is  bisected  by  0.  Therefore  all  the  diagonals 
pass  through  0  and  aie  bisected  by  that  point.     Q.E.D. 

v<j«^.  j^cf.  iiit3  \){jiii.i  O  iiirougn  wnicn  all  tiie 
diagonals  pass  is  called  the  centre  of  the  parallelo- 
piped. 


I 


i 


822 


BOOK  JX.    OF  POL  TUEDRONS. 


m 


'fry' 


B 


Theorem  VI. 

690.  The  sum  of  the  squares   upon  the  four 
diagonals  of  a  parallelopiped  ^ 

is  equal  to   the  sum   of  the  ^ 
squares  upon  its  twelve  edges. 

Hypothesis.     ABGD  -  BFOII, 
any  parallelopiped. 

Conclusion.   AH'-\-BQ*-^CF' 
4-  DE'  =  ^(A  C  +  AB'  4-  AE'). 

Proof.     Draw  the  diagonals  of  (j 
any  pair  of  opposite  faces,  as  ^Z> 
and  BO,  EH  and  FO.     Then— 

1.  AH  and  DE  are  diagonals  of  the  parallelogram  AD  HE 
(§  688,  1).     Therefore 

AH' -^  DE' =  2AD' -{- 2AE\  (§316) 

2.  In  the  same  way, 

BG'  +  CF'  =  2BC*  +  2BF* 
=  2BC'  -f  2AE\ 

3.  Adding  (1)  and  (2), 

AH'  +  BO'  +  OF'  +  DE'  =  2(AD'  +  BC^)  +  4^JSr«. 

4.  Because  AD  and  ^C  are  dis'.gonals  of  the  parallelo- 
gram ABGD, 

AD'  +  BC  =  2AB'  +  2A0\  (§316) 

6.  Substituting  this  result  in  (3),  we  have 
Sum  of  squares  of  diagonals  =  4AB'  -j-4:AC*  -\-  4:AE*. 
6.  Since  there  are  four  edges  equal  to  AB,  four  equal  to 
AC,  and  four  equal  to  AE,  this  sum  is  equal  to  the  sum  of 
the  squares  of  all  the  edges,  and 

Sum  of  squares  of  diagonals  =  sum  of  squares  of  edges. 

Q.E.D. 

Theorem  VII. 

691.  The  four  diagonals  of  a  rectangular  paral- 
lelopiped are  equal  to  each  other. 

Hypothesis.   ABCD-EFGH,  a  rectangular  parallelopiped. 
Conclusion.    The  diagonals  AH,  BG,  CF,  and  DE  Skre  all 
equal. 


PjiRALLELOPIPEDS. 


823 


\e  four 


-"-^ 


B 
iADHE 

(§316) 


\AE\ 
>arallelo- 

(§316) 

AE\ 
equal  to 
I  sum  of 

dges. 
Q.E.D. 


paral- 

lopiped. 
E  are  all 


Proof.  1.  Because  the  faces  AFvLnd  AG  are  each  at  right 
angles  to  the  fade  AD  (§  G85),  their  lino  of  intersection 
AE  18  also  perpendicular  to  that  „ 

face  (§  633),  and  to  the  line  ^Z>  in  0 
that  face  (§  584). 

2.  Therefore  ADEH  is  a  rect- 
angle, and  its  diagonals  AH  and 
DE  are  equal. 

3.  It  may  be  shown  in  the  same 
way  that  any  other  two  diagonals 
are  equal.  Therefore  these  diago- 
nals are  all  equal  to  each  other. 

Q.E.D. 

Theorem  VIII. 

693.  The  square  of  each  diagonal  of  a  reetangu^ 
lar  parallelopiped  is  equal  to  the  sum  of  the  squares 
of  the  three  edges  which  meet  at  any  vertex. 
Hypothesis.     Same  as  in  Theorem  VII. 
Conclusion.    AH''  =  AB^  -{■  AC^  •{-  AE\ 
Proof.     1.  Because  AEH  is  a  right  angle  (Th.  VII.,  1), 
AH'  =  AE'  +  EH* 
=  AE'  +  AI)\ 

2.  Because  ABD  is  a  right  angle, 

AD'  =  AB'  +  BD' 
=  AB'-\-AC\ 

3.  Comparing  with  (1), 

AH'  =  AE'  ■^AB'-\-A  G\     Q.E.D. 

693.  Scholium.  This  theorem  might  have  been  regarded 
as  a  corollary  from  the  two  preceding  ones.  But  we  have 
preferred  an  indejjendent  demonstration,  owing  to  its  impor- 
tance. It  may  be  considered  as  an  extension  of  the  Pytha- 
gorean proposition  from  a  plane  to  space. 

Pyramids. 

694.  Def.  A  Dvramid  is  a  Dolvhedron  of  whicli 
all  the  faces  except  one  meet  in  a  point. 

The  point  of  meeting  is  called  the  vertex. 


■  * 


(    »   I 


^^^^Wp 


324 


BOOK  IX.    OF  P0LTHEDR0N8. 


Remark,  The  face  which  does  not  pass  through  the 
vertex  is  taken  as  the  base. 

696.  Def.  The  faces  and  edges 
which  meet  at  the  vertex  are  called 
lateral  faces  and  edges. 

696.  JDef.  The   altitude   of   a 

pyramid  is  the  perpendicular  dis- 
tance from  its  vertex  to  the  plane 
of  its  base. 

6911.  Def.   A  pyramid  is  said  to        a  pyramid, 
be   triangular,  quadrangular,   pentagonal,  etc.,  ac- 
cording as  its  base  is  a  triangle,  a  quadrilateral,  a 
pentagon,  etc. 

698.  'Def.  If  the  vertex  of  a  pyramid  is  cut  off  by 
a  plane  parallel  to  the  base,  that  part  v^hich  remains 
is  called  a  frustum  of  a  pyramid. 

Theorem  IX. 

699.  If  a  pyramid  be  cut  hy  a  plane  parallel  to 
the  base,  then — 

I.  The  edges  and  the  altitude  are  similarly  di- 
vided. 

II .  The  section  is  similar  to  the  base. 

Hypothesis.     0-ABCDE,  a  pyramid;  OP,  its  altitude; 
dbcde,  a  section  of  the  pyramid  by  a 
plane  parallel  to  the  base  ABODE, 
cutting  the  altitude  line  at  g. 

Conclusions. 
I.   OF  :Og::OB:  Ob  ::OC:Oc,  etc. 

II.  The  polygon  abcde  is  similar  to 
the  polygon  ABODE.  ^ 

Proof.     1.  Because  AB  and  ah  are 
the    intersections    of    parallel  planes    [/ 
with  the  plane  OAB, 

ah  II  AB.  (§614) 


PYRAMIDS. 


825 


0 


>c 


2.  Therefore  the  triangles  OAB  and  Odb  ar^  simUar,  and 
OA  :  Oa  ::  OB  :  Ob 


bB. 


whence  Oa   -.  aA  :-.  Ob 

3.  In  the  same  way  it  may  be  shown  that  00,  OP,  etc 
are  all  divided  similarly  at  c,  g,  etc.    Q.E.D.  *' 

4.  It  may  also  be  shown,  as  in  (1),  that  ea^h  side  of  the 

polygon  abcde  is  parallel  to  the  corresponding  side  of  ABODE 

Therefore  the  angles  of  the  two  polygons  are  respectively 

equal  (§  608),  and  the  polygons  are  equiangular  to  each  other. 

6.  Because  the  bases  ab  and  AB  of  the  triangles  Oab  and 
OAB  axe  parallel, 

_  AB  :ab  =  OA  :0a. 

In  the  same  way, 

BO:  be::  OB:  Ob  ::  OA  :  Oa, 
CD:cd::  00  :  Oc  ::  OA  :  Oa, 
etc.  etc. 

6.  Therefore  the  polygons  ABCDE  and  abcde,  having 
their  angles  equal  and  the  sides  containing  the  equal  angles 
proportional,  are  similar  to  each  other.     Q.E.D. 

Corollary  1.  Because  abcde  and  ABCDE  ^ve  similar  poly- 
gons, their  areas  are  as  the  squares  of  ab  and  AB:  that  is,  as 
Oa'  :  0A\  or  Og'  :  OP'  (§435).     Hence: 

700.  ITie  areas  of  parallel  sections  of  a  pyramid  are  pro- 
portional  to  the  squares  of  the  distances  of  the  cutting  planes 
from  the  vertex. 

Cor.  2.  The  base  of  a  pyramid  may  be  regarded  as  one  of 
the  plane  sections,  so  that  if  two  pyramids  have  equal  bases 
and  altitudes,  the  plane  sections  made  by  the  bases  are  equal, 
and  the  distances  of  these  sections  from  the  vertices  are  also 
equal.     Hence: 

701.  In  pyramids  of  equal  bases  and  altitudes,  parallel 
plane  sections  at  equal  distances  from  the  vertices  are  equal  in 
area,  ^ 


m\\ 


'i  I 

lit 


'4 


326 


BOOK  IX,    OF  P0LTHEDB0N8. 


CHAPTER    II. 

THE    FIVE    REGULAR    SOLIDS. 


703.  Def.  A  regular  polyhedron  is  one  of  which 
all  the  faces  are  identically  equal  regular  polygons 
and  all  the  polyhedral  angles  are  identically  equal. 

Remaek.  a  regular  polyhedron  is  familiarly  called  a 
regular  solid. 

703.  Problem.  Tofind>  Jiow  many  regular  solids 
are  possible. 

1.  Let  us  consider  any  vertex  of  a  regular  polyhedron. 
Since  at  least  three  faces  must  meet  to  form  the  polyhedral 
angle  at  each  vertex,  and  since  the  sum  of  all  the  plane  angles 
which  make  up  the  polyhedral  angle  must  be  less  than  360° 
(§  661),  we  conclude: 

Each  angle  of  the  faces  of  a  regular  solid  must  be  less 

than  120°. 

2.  Since  the  angles  of  a  regular  hexagon  are  each  120°, 
and  the  angles  of  every  polygon  of  more  than  six  sides  yet 

greater,  we  conclude: 

Bach  face  of  a  regular  solid  must  have  less  than  six  sides. 
Such  faces  must  therefore  he  either  triangles,  squares,  or 

pentagons. 

3.  If  we  choose  equilateral  triangles,  each  polyhedral  angle 
may  have  either  3,  4,  or  5  faces,  because  3  X  60°,  4  x  60°, 
and  5  X  60°  are  all  less  than  360°.  It  cannot  have  6  faces, 
because  each  angle  of  the  triangle  being  60°,  six  angles  would 
make  360°,  reaching  the  limit.  Therefore  no  more  than 
three  regular  solids  may  be  constructed  with  triangles. 


4.  Because  each  angle  of  a  square  is 

Therefore  only  one  regular 
6.  Because  each  angle 


90°,  three  is  the  only 

■nnlvViAflral    an  or]  ft. 

have  square  faces. 

alar  pentagon  is  108°,  a 


which 
ilygons 
ual. 
called  a 


solids 

^hedron. 
lyhedral 
e  angles 
lan  360° 

f  le  less 

3h  120°, 
jides  yet 

ix  sides. 
',ares,  or 

ral  angle 
4  X  60°, 
)  6  faces, 
es  would 
3re  than 

IS. 

the  only 

rl    ftTijylfi. 

~    — o — 

!S. 

3  108°,  a 


BEOULAR  80LID8.  '  ^^l 

polyhedral  angle  cannot  be  formed  of  more  than  +}ir«n  >.    . 

..J'  "^^  *^«^?;f?^«  conclude  that  not  more  than  five  re^kr 

e JSt^:  ''^^^TTsJ'z  "^  ^-^^  *^"^  «^"*'^'- 

CO^,  and  join  them  at  the  common 
vertex,  0,  ABO  win  be  an  identi- 
cally equal  equilateral  triangle. 
Therefore  a  polyhedron  will  be 
formed  having  as  faces  four  identi- 
cal equilateral  triangles. 

This  solid  is  a  regular  tetra- 
hedron. 

lelop'Jedof'l;^^^^^^^^^^^ 

is  clear  from  §§  685-693  ^    '''*    ^*'  construction 

T06.  The  Octahedron.  Let 
-45CZ>  be  a  square;  0,  its  centre; 
^{f,  a  Ime  passing  through  0  perpen- 
dicular to  the  plane  of  the  square. 

On  this  line  take  the  points  P  and  A«=££---'-/-Or'-±i::^-^D 
Q,  such  that  PA,  PB,  PO,  and  Pi> 
also  QA,  QB,  QO,  and  QB,  are  each 
equal  to  a  side  of  the  square.     The 
^f^^f'^BOD^Q  will  be  a  regular 

PrlTT  1''  ^''  '^^'^  ''^^^  ^^"''J^teral  triangles 
Pm/.  1.  Because  all  the  lines  from  Por  0  to  7*  P  n      a 
D  are  equal,  all  the  eight  triangles  which  form  fh.^'    ' 
equal  and  equilateral.  °^  *^®  ^*^^s  are 

3.  Let  a  diagonal  be  drawn  from  A  to  D 

Because  0  is  the  centre  of  the  square  jpnn  +t.-    ^• 
onal  will  y>oc,«  fi, 1    ^        ,     .     «4"'*re  ^//6x',  this  diair- 

^,  ^,  ^,  and  D  are  in  one  plane. 


328 


BOOK  IX.    OF  POL  THEDR0N8. 


3.  In  the  triangles  APDy  ABD,  AQD  we  have 

Side  AD  common, 

All  the  other  sides  equal. 
Therefore  the  triangles  are  identically  equal;  and  because 
ABD  is  a  right  angle,  APD  and  AQD  are  also  right  angles, 
and  the  quadrilateral  APDQ  is  a  square  equal  to  the  square 

ABCD.  .      ^        ^ 

4.  Because  the  polyhedral  angle  at  B  has  its  four  faces 
and  its  base  APDQ  equal  to  the  four  faces  and  the  base 
ABCD  of  the  polyhedral  angle  at  P, 

Polyhedral  angle  B  =  polyhedral  angle  P. 

5.  In  the  same  way  it  may  be  shown  that  any  other  two 
polyhedral  angles  are  equal.  Therefore  the  figure  P-ABGD-Q 
is  a  regular  solid  having  eight  equilateral  triangles  for  its  faces. 

This  solid  is  called  the  regular  octahedron. 

707.  The  Dodecahedron.   Taking  the  regular  pentagon 
ABODE  as  a  base,  join  to  its  sides 
those  of  five  other  equal  pentagons, 
so  as  to  form  five  trihedral  angles 
at  A,  B,  0,  D,  and  E,  respectively. 

Because  the  face  angles  of  these 
trihedral  angles  are  equal,  the  angles 
themselves  are  identically  equal. 

(§  660) 

Therefore  the  dihedral  angles 
formed  along  the  edges  AK,  BL, 
CM,  etc.,  are  equal  to  the  dihedral  angles  AB,  BO,  etc. 

Therefore  the  face  angles  EAK,  LBC,  etc.,  are  identi- 
cally equal  to  the  angles  of  the  original  regular  pentagons. 

Pass  planes  through  JT^Pand  OKF,  etc.,  and  let  FP  be 
their  line  of  intersection.  Then,  continuing  the  same  course 
of  reasoning,  it  may  be  shown  that  the  face  angles  GKF, 
FLJ,  etc.,  are  all  angles  of  108°,  or  those  of  a  regular  penta- 
gon. Completing  this  second  series  of  five  pentagons,  we  shall 
have  left  a  pentagonal  opening,  which  being  closed,  the  surface 

Ol   LU6   UUi Y UUtliAJli  will   J7C  tJUiiipicluu.  tto  DiiUTTii  111   tuo  Uxaijji tiixii. 

The  solid  thus  formed  has  \%  pentagons  for  its  sides,  and 
is  called  a  regular  dodecahedron. 


BEQULAB  80LID8. 


829 


because 
angles, 
square 

ir  faces 
he  base 


her  two 
BCD'Q 
is  faces. 

entagou 


stc. 

)  identi- 
yons. 
\.FP  be 
le  course 
js  GKF, 
ir  penta- 
we  shall 
e  surface 

XiOigX  CtXi.i« 

deS;  and 


o  J^^\  ^^^  IcosaTiedron.  Let  five  equilateral  tnauffiee  form 
a  polyhedral  angle  at  P,  such  that  ^laugies  form 

the  dihedral  angles  along  FA,  FB 
etc.,  shall  all  bo  equal. 

The  base  ABCDB  will  then  form 
a  regular  pentagon. 

Complete  the  polyhedral  angles 
at  A,  B,  O,  D,  and  B  by  adding  to 
each  three  other  equilateral  triangles, 
and  making  the  dihedral  angles 
around^,  ^,C;  etc.,  all  equal. 

It  can  be  then  shown,  as  in  the 
case  of  the  dodecahedron,  that  the  lines  F  G  H  T  1  ^m 
form  a  second  regular  pentagon.  '     '     '  ^'  "^  ^^^ 

This  solid  is  called  the  icosahedron. 
709.  The  five  regular  solids  are  therefore- 

The  tetrahedron,  formed  of  3  triangles. 

Ihe  cube,  or  hexahedron,  formed  of   6  squares. 

The  octahedron,  formed  of  8  triangles. 

The  dodecahedron,  formed  of  13  pentagons. 

The  icosahedron,  formed  of  20  triangles. 

Theoeem  X 

TIO.  The  perpendiculars  through  the  centres  of 
the  faces  of  a  regular  solid  meet  in  a  point  which 

lyj^^^^fstantfr^  all  the  faces,  f%o:i  aUhe 
edges,  and  from  all  the  vertices. 

lar  fj.fr'-    ^f^^f.  ^^?  ^^^^^^  *^«  ^^«««  0^  a  regu- 
lar polyhedron,  intersecting  along  the  edge  ^5-  0   0    thP 

centres  of  these  faces;  OR.  OR.  ^.rr..^^^^^.t,: .yl  1\  *^' 
Conclusion,      These   perpendiculars  meet  all  the  perpen- 
diculars through  the  centres  of  the  other  faces  in  a  point 
R  equally  distant  from  all  the  faces,  edges,  and  vertices 


J  i  1 


330 


BOOK  IX.    OF  POL  THEDR0N8, 


Proof,   From  0  and  Q  drop  perpendiculars  upon  the  edge 
AB,    Then— 

1.  Because  these  perpendic- 
ulars are  dropped  from  the  cen- 
tres of  regular  polygons,  they 
will  fall  upon  the  middle  point 
P  of  the  common  side  AB, 

2.  Because  PO  and  PQ  are 
perpendicular  io  AB  oi  the  same 
point  Py  and  OR  and  QR  arc 
perpendiculars  to  the  intersecting 
planes,  they  will  meet  in  a  point  (§  6?.7). 

Let  R  be  their  point  of  meeting.    Join  PR.    Then — 

3.  In  the  right-angled  triangles  POR  and  PQR  we  have 
Side  PR  common, 

pd  =  PQ  (being  apothegms  of  equal  polygons). 
Therefore  these  triangles  are  identically  equal,  and 

OR  =  QR. 
Angle  PRO  =  angle  PRQ. 

4.  If  S  be  the  centre  of  any  other  face  adjacent  to 
ABODE,  it  can  be  shown  in  the  same  way  that  the  perpen- 
dicular from  S  will  meet  QR  in  a  point. 

5.  Because  the  angles  between  the  faces  Q  and  8  are  the 
same  as  between  0  and  Q,  it  may  be  shown  that  the  perpen- 
dicular from  8  will  meet  QR  in  the  point  R. 

6.  Continuing  the  reasoning,  it  will  appear  that  all  the 
perpendiculars  from  the  centres  of  the  faces  meet  in  the  same 

point  R. 

7.  If  from  R  perpendiculars  be  dropped  upon  all  the  edges 
and  all  the  vortices,  these  perpendiculars,  together  with  those 
upon  the  corresponding  faces  and  the  lines  like  ^P  and  QB 
from  the  centres  of  the  faces  to  the  edges  and  vertices,  will 
form  identically  equal  triangles.  Hence  will  follow  the  con- 
clusion to  be  demonstrated. 

Note.  We  have  given  but  a  brief  outline  of  the  demonstration, 
which  the  student  may  complete  as  an  exercise.  The  conclusions  may 
also  be  considered  as  following  immediately  from  the  symmetry  of  the 
polyhedron. 


the  edge 


len — 
we  haye 

s)., 
id 


acent  to 
e  perpen- 

6^  are  the 
e  perpen- 

it  all  the 
the  same 

the  edges 
i^ith  those 
^  and  QB 
tices,  will 
'  the  con- 

lonstration, 
asions  may 
letry  of  the 


BBGULAB  SOLIDS. 


331 


Theorem  XI. 

711.  ^  regular  solid  is  symmetrical  with  respect 
to  all  its/aces,  edges,  and  vertices. 

Proof,     Let  ABO  be  a  face  of  any  regular  polyhedron- 
A,  B,  and  G  will  then  be  yer-  ' 

tices. 

Let  AD,  BE,  BF,  etc.,be 
the  edges  going  out  from  these 
yertices. 

Now  moye  the  polyhedron  so 
as  to  bring  any  other  face  into 
the  position  ABC.  This  can  be 
done,  because  the  faces  are  all 
identically  equal. 

Because  the  polyhedral  angles 
are  all  identically  equal,  whatever  polyhedral  angles  take  the 
positions  A,  B,  0,  their  faces  and  edges  will  coincide  with  the 
positions  of  the  faces  and  edges  already  marked  in  the  figure. 

Because  the  edges  are  all  of  equal  length,  the  yertices  at 
the  ends  of  D,  E,  F,  0  will  fall  into  the  same  positions  where 
the  former  yertices  were. 

Continuing  the  reasoning,  the  whole  polyhedron  will  be 
found  to  occupy  the  same  space  as  before,  eyery  face,  edge, 
and  yertex  falling  where  some  other  face,  edge,  or  yertex  was 
at  first. 

Because  this  is  true  in  whatever  way  the  positions  of  the 
faces  may  be  interchanged,  the  polyhedron  is  symmetrical. 

Q.E.I). 

713.  Corollary.  Conyersely,  if  a  polyhedron  le  such 
that,  when  any  one  face  is  brought  into  coincidence  with  the 
position  of  any  other,  every  other  face  shall  coincide  with  the 
former  position  of  some  face,  the  polyhedron  is  regular. 

Theorem  XII. 
If  a  plane  oe  passed  tlirougTi  each  vertex  of 
-  solid,  at  right  angles  to  the  radius,  these 
planes  will  he  the  faces  of  another  regular  solid 


a 


regul 


ar 


n\ 


5-!| 


332 


BOOK  IX.    OF  POLTHEDRONa. 


Proof,    1.  Let  A,  B,  and  (7  be  any  vertices  of  a  regular 
solid,  and   0   its   centre.      Imagine 
planes  passed  through  4>  B,  and  G, 
perpendicular  to   OA,  OB,  and  00 
respectively,  and  cut  off  along  their  A.-^ 
lines  of  intersection,  so  as  to  form       \,  /         y:^B 

the  faces  of  another  solid.     We  call 
this  the  new  solid,  and  the  original  '^^"'      ^ 

one  the  inner  solid, 

2.  Because  the  inner  solid  is  regular,  if  we  bring  any  other 
of  its  faces  into  the  position  ABC,  the  whole  solid  will  occupy 
the  same  position  a?  before  (§  711). 

3.  Because  each  face  of  the  new  solid  is  at  right  angles 
to  the  end  of  a  radius  to  some  vertex  of  the  inner  solid, 
f.nd  these  radii  all  coincide  with  the  former  positions,  the 
plane  of  each  face  of  the  new  solid  will,  when  the  change  of 
position  is  made,  take  the  position  of  the  plane  of  some  other 

face. 

4.  Therefore  the  edges  in  which  these  planes  intersect 

will  take  the  positions  of  other  edges. 

5.  Therefore  the  vertices  where  these  edges  meet  will  take 
the  positions  of  other  vertices. 

6.  Therefore  the  new  solid  occupies  the  same  space  as 
before  the  change,  and  is  consequently  symmetrical  with 
respect  to  all  its  faces,  edges,  and  vertices. 

Therefore  it  is  a  regular  solid  (§  712).     Q.E.D. 

714.  Def.  A  pair  of  polyhedrons  such  that  each 
face  of  the  one  corresponds  to  a  vertex  of  the  other 
are  called  sympolar  polyhedrons. 

Theobem  XIII. 

716.  Every  regular  solid  has  as  many  faces  as 
its  sympolar  has  wrtices,  and  as  many  edges  as  its 
sympolar  has  edges. 

Proof,  1.  Continuing  the  reasoning  of  the  last  theorem, 
it  is  evident  that  the  centres  of  the  faces  of  the  new  solid 
coincide  with  the  vertices  of  its  sympolar. 


RBOULAR  SOLIDS. 


838 


'egular 


y  other 
occupy 

angles 
p  solid, 
us,  the 
ange  of 
le  other 

atersect 


2.  Because  each  edge  of  a  regular  solid  is  equally  distant 
from  the  centres  of  two  adjoining  faces,  each  edge  of  the  new 
solid  will  be  equally  distant  from  two  adjoining  vertices  of  the 
sympolar. 

3.  Since  every  two  such  vertices  are  connected  by  an  edge 
of  the  sympolar,  the  new  soHd  will  have  as  many  edges  as  the 
sympolar,  each  edge  of  the  one  being 
at  right  angles  and  above  the  edge  of 
the  sympolar.     Q.E.D. 

Let  Ay  B,  and  C  be  three  vertices  ^-      ^        "^        -'* 
of  the  sympolar,   and    therefore    the 
centres  of  three  faces  of  the  new  solid. 

4.  By  what  has  just  been  shown,  P«, 
Ph,  and  Pc  will  be  three  edges  of  the 
new  solid,  meeting  in  a  vertex  at  P, 
Therefore — 

5.  The  new  solid  will  have  a  vertex  over  the  centre  of  each 
side  of  the  sympolar,  and  so  will  have  as  many  vertices  as  the 
sympolar  has  faces.    Q.E.D. 


I 

0 


rill  take 


pace  as 
al  with 


at  each 
e  other 


aces  as 
s  as  its 


theorem, 
lew  solid 


Theorem  XIV. 

716.  The  sympolar  of  a  polyhedron  whose  faces 
have  8  sides  will  have  S-hedral  angles. 

Proof.  The  vertex  of  one  polyhedron  being  at  P  over  the 
centre  of  the  face  of  its  sympolar,  the  edges  meeting  at  this 
vertex  are  each  perpendicular  to  an  edge  of  the  face  of  the 
sympolar. 

Therefore  if  the  face  has  S  sides,  the  polyhedral  angle 
above  it  will  have  S  edges,  and  therefore  8  faces.     Q,E.D. 

•717.  Corollary.  Conversely,  the  sympolar  of  a  poly- 
hedron, whose  vertices  are  S-hedral  will  have  S-sided  faces. 

718.  Corollary.  What  pairs  of  regular  solids  are  sympolar 
to  each  other  can  be  readily  determined  from  the  preceding 
theorems. 

The  tetrahedron  has  four  vertices.  Therefore  its  sym- 
polar has  four  faces,  and  is  therefore  another  tetrahedron. 


334 


BOOK  IX.    OF  POLTHEDROm. 


The  cube  has  8  trihedral  vertices  and  6  four-sided  faces. 
Therefore  its  polar  has  8  triangular  faces  and  6  four-hedral 
vortices.     This  is  the  octahedron. 

Conversely,  the  sympolar  of  the  octahedron  is  the  cube. 
They  each  have  8  edges. 

The  dodecahedron  has  12  pentagonal  faces  and  20  tri- 
hedral vertices.  Therefore  its  sympolar  has  12  five-hedral 
vertices  and  20  triangular  faces.     This  is  the  icosahedron. 

Each  of  these  polyhedrons  has  30  edges. 

These  results  are  shown  in  the  following  table,  where  the 
headings  at  the  top  of  each  column  refer  to  the  solid  on  the 
left,  and  those  at  the  bottom  to  its  sympolar  on  the  right. 


SoUd. 

i 

Number 

of  sides 

to  each 

face. 

Number 
of  faces. 

Number 
of  edges. 

Number 

of 
vertices. 

Number 

of  edges 

at  each 

vertex 

Tetrahedron. 

Cube. 

Octahedron. 

Dodecahedron. 

Icosahedron. 

8 
4 
8 
6 
8 

4 
6 

8 

n 

20 

6 

13 

n 

80 
80 

4 
8 
6 

ao 

12 

8 
8 
4 
8 
6 

Tetrahedron. 

Octahedron. 

Cube. 

Icosahedron. 

Dodecahedron. 

Number 

of  edges 

at  each 

vertex. 

Number 

of 
vertices. 

Number 
of  edges. 

Number 
of  faces. 

Number 

of  sides 

to  each 

face. 

Sjrmpolar 

Note  that  each  column  applies  to  two  solids.  For  instance,  the  left- 
hand  column  shows  the  number  of  sides  to  each  face  of  the  solid  named 
at  the  leit,  and  the  number  of  edges  at  each  vertex  of  the  £olid  named  at 
the  right. 


i  faces, 
-hedral 

e  cube. 

20  tri- 
)-hedral 
ron. 

lere  the 
on  the 
ght. 


ihedron. 
hedron. 

5. 

ihedroo. 
)cahedron. 


mpolar 
solid. 


,  the  left- 
id  named 
named  at 


BOOK  X. 
OF    CURVED    SURFACES. 


CHAPTER     I. 

THE  SPHERE. 

Definitions. 

719.  Def.  A  curved  surface  is  a  surface  no  part 
of  which  is  plane. 

■720.  Def.  A  spherical  surface  is  a  surface  which 
is  everywhere  equally  distant  from  a  point  within  it 
called  the  centre. 

•731.  Bef.  A  sphere  is  a  solid  bounded  by  a 
spherical  surface. 

Note  1.  A  spherical  surface  may  also  be  described  as  the 
locus  of  the  point  at  a  given  distance  from  a  fixed  point 
called  the  centre. 

Note  2.  In  the  higher  geometry  a  spherical  surface  is 
called  a  sphere.  We  shall  use  this  appellation  when  no  con- 
fusion will  thus  arise. 

732.  Def.  The  radius  of  a  sphere  is  the  distance 
of  each  point  of  the  surface  from  the  centre. 

733.  Def.  A  diameter  of  a  sphere  is  a  straight 
line  passing  through  its  centre,  and  terminated  at  both 
ends  by  the  surface. 

Corollary.  Every  diameter  is  twice  the  radius;  therefore 
all  diameters  of  the  same  sphere  are  equal. 


•^QA. 


V 


-a-       -'-■   ■-.-   w-rw  m  wf        K_5.i.«iiiff^      I;t   t    n.    >-l  ; 


v% ; 


—  1. 


iSjut?  lA^  a,  D^^iici  c  its  u,  pia-ue 
which  has  one  point,  and  one  only,  in  common  with 
the  sphere. 


1 


IB 


336 


BOOK  X    OF  CURVED  SURFACEB. 


726.  Def.  A  line  is  tangent  to  a  sphere  when  it 
touches  the  spherical  surface  at  a  single  point. 

i       726.  Def.  Two  spheres  are  tangent  to  each  other 
when  they  have  a  single  point  in  common. 

727.  Def.  A  section  of  a  sphere  is  the  curve 
line  in  which  any  other  surface  intersects  the  spherical 
surface. 

728.  Def  Opposite  points  of  a  sphere  are  points 
at  the  ends  of  a  diameter. 

Theorem  I. 

729.  Etiery  section  of  a  sphere  hy  a  plane  is  a 
circle  of  which  the  centre  is  the  foot  of  the  perpen- 
dicular f  rem  the  centre  of  the  sphere  upon  the  plane. 

Hypothesis.    AB,  any  sphere;  0,  its  centre;  QRS,  the 
curve  in  which  a  plane  intersects  the 
spherical  surface;  OC,  the  perpen- 
dicular from   0  upon  the  cutting 
plane. 

Conclusion.     QRS  is    a   cirde^l 
having  G  as  its  centre. 

Proof.  1.  Because  the  lines 
OQ,  OR,  OS&ve  radii  of  the  sphere, 
they  are  equal. 

2.  Because  they  are  equal,  they  meet  the  plane  QRS  at 
equal  distances  from  the  foot  G  of  the  perpendicular  (§  595). 

Therefore  the  curve  QRS  is  a  circle  around  (7  as  a  centre. 

Q.E.D. 

730.  Corollary.  The  line  through  the  centre  of  a  circle 
of  the  sphere,  perpendicular  to  its  plane,  passes  through  the 
centre  of  the  sphere. 

731.  Def.  The  circular  section  of  a  sphere  by  a 
plane  is  called  a  circle  of  the  sphere. 

732.  Def.  If  the  cutting  plane  passes  through  the 
centre  of  the  sphere,  the  circle  of  intersection  is  called 


THE  BPUERB. 


837 


tien  it 

other 

curve 
lerical 

points 


e  ts  a 
erpen- 
plane. 

W,  the 


QRS  a.t 

§  595). 
i  centre. 
.E.D. 

■  a  circle 
mgh  the 

:e  by  a 


ign  rne 
3  called 


^?'l''l^*?''!\®  ""^x.**"*  "P^*"'  ^"^  *^«  *^o  parts  into 
which  It  divides  the  sphere  are  ciilled  hemiipherei. 

733.  Corollary.     All  great  circles  of  the  sphere  are  equal 
to  each  other.  ^  x^  i^  «•!*» 

xt,  "^^l*  ^y-  "^^  ^"^^  P^^'^*^  ^^  w^ich  a  perpendicular 
through  the  centre  of  a  circle  of  the  sphere  intersects 
the  surface  of  the  sphere  are  called  poles  of  the  circle. 

Cor.  2.  Because  the  perpendicular   passes    through    the 
centre  of  the  sphere: 

736.  The  two  poles  of  every  circle  of  the  sphere  are  at  the 
extremities  of  a  diameter,  and  so  are  opposite  points  (§  728). 

Theorem  II. 

736.  Every  great  circle  divides  the  sphere  into  two 
taentically  equal  hemispheres. 

Hypothesis.  AB,  a  circle  of  the  sphere,  having  the  centre 
0  of  the  sphere  in  its  plane  and 
dividing  the  sphere  into  the  parts 
M  and  iV; 

Conclusion.   The  parts  Jfand  JV 
are  identically  equal. 

Proof.     Turn  the  part  M,  on  0>| 
as  a  pivot  so  that  the  plane  ^^  shall 
return  to  its  own  position  but  be 
inverted.     Then— 

1.  Because  the  centre  0  remains 
fixed,  the  great  circle  AB  will  fall 
upon  its  own  trace. 

2.  Because  the  surfaces  if  and  JV  are  everywhere  equally 
distant  from  the  centre,  they  will  coincide  throughout. 

Therefore  the  parts  are  equal.     Q.E.D. 

Theorem  III. 
737.  Any  two  great  circles  Used  each  other. 
Proof.  Let  AB  and  CD  be  the  two  great  circles.   Because 
the  planes  of  these  circles  both  pass  through  the  centre  of 


"-ii 


r  I 


338 


BOOK  X     OF  OUBVBD  SUBFACES. 


the  sphere,  their  line  of  intersection  is  a  diameter  of  the 
sphere,  and  therefore  a  diameter  of  each  circle. 

Hence  it  divides  each  circle  into  two  equal  parts.    Q.E.D. 

738.  Corollary.  If  any  number  of  great  circles  pass 
through  a  point,  they  will  also  pass  through  the  opposite 
point. 

Theorem  IV. 

739.  Through  three  points  on  a  sphere  one  circle, 
and  only  one,  can  be  passed. 

Proof.  1.  Three  points  determine  the  position  of  a  plane 
passing  through  them  (§  580). 

2.  This  plane  cuts  the  sphere  in  a  circle  of  the  sphere. 

(§  'J'29) 

3.  Because  the  three  points  lie  both  upon  the  sphere  and 
in  this  plane,  they  lie  in  thi^  circle. 

4.  Only  one  circle  can  pass  through  these  points  (§  241). 
6.  Therefore  this  circle,  and  no  other,  passes  through  the 

three  points.    Q.E.D. 

Theorem  V. 

740.  77:  rough  two  points  upon  a  sphere,  not  at 
the  extremities  of  a  diameter,  one  great  circle,  and 
only  one,  can  pass. 

Proof.  1.  Because  the  plane  of  the  great  circle  must  pass 
through  the  centre  of  the  sphere,  the  centre  and  the  two 
points  on  the  surface  determine  its  position  (§  580). 

2.  But  if  the  points  are  at  the  extremities  of  a  diameter, 
the  centre  is  in  the  same  straight  line  with  them,  and  an 
infinite  number  of  planes  may  pass  through  them  (§  577,  II.). 

741.  Def.  The  arc  between  two  points  on  a  sphere 
means  the  arc  of  the  great  circle  passing  through 
these  points. 

743.  Equal  arcs  upon  the  same  sphere  subtend 
equal  angles  at  the  centre. 


TBB  SPHERE. 


839 


Proof,  Because  all  great  circles  are  equal,  their  arcs  are 
arcs  of  equal  circles  (§  733). 

Because  their  centres  are  in  the  centre  of  the  spheres,  and 
because  equal  arcs  subtend  equal  angles  at  the  centre  (§  308) 
their  equal  arcs  subtend  equal  angles  at  the  centre  of  the 
sphere.     Q.E.D. 

743.  Corollary,  Equal  chords  in  the  sphere  suMend 
equal  angles  at  the  centre. 

744.  Corollary.  The  angular  distance  letween  two  points 
on  the  sphere  may  le  measured  either  by  the  great  circle  join- 
ing them,  or  by  the  angle  between  the  radii  drawn  to  them. 

Note.    By  the  distance  of  two  such  points  is  c<    imonly  meant  their 
angular  distance 

Theorem  VII. 

745.  All  points  of  a  circle  of  the  sphere  are 
equally  distant  from  a  pole. 

Hypothesis.  QR,  a  circle  of  the  sphere  AB-,  P,  P',  the 
poles  of  the  circle. 

Conclusion.  Every  point  of  the 
circle  QR  is  equally  distant  from 
P  and  equally  distant  from  P'.  q^ 

Proof.  1.  Because  PP'  is  a 
line  through  the  centre  of  the  a  I 
circle  perpendicular  to  its  plane, 
every  point  of  this  line  is  equally 
distant  from  all  points  of  the 
circle  (§  594). 

3.  Therefore  P  and  P',  being  on  ^ 

the  line,  are  each  equally  distant  from  all  points  of  the  circle. 

Q.E.D. 

746.  Corollary.  Every  point  of  a  circle  of  the  sphere  is 
at  an  equal  angular  distance  from  the  pole. 

747.  Bef.  The  polar  distance  of  a  circle  of  the 
sphere  is  the  common  angular  distance  of  all  its 
points  from  either  pole. 


'sU 


^ffii 


'W\ 


340 


BOOKX.    OF  OUBVED  SURFACES. 


Cor,  2.  The  angular  distance  of  two  poles  is  a  semicircle, 
because  they  are  at  the  extremities  of  a  diameter.     Hence: 

748.  The  sum  of  the  distances  of  a  circle  from  its  two 
poles  is  a  semicircle. 

749.  Cor.  3.  Every  point  of  a  great  circle  of  the  sphere 
is  half  a  semicircle  or  a  right  angle  distant  from  each  pole. 

750.  Def.  A  quadrant  is  an  arc  of  one  fourth  a 
great  circle,  or  half  a  semicircle. 

Corollary.  A  quadrant  subtends  a  right  angle  from  the 
centre  of  the  sphere. 

Illustration.  If  AB  is  a  great 
circle  of  the  sphere,  and  P  and  P' 
its  poles,  then 

ArcP^P'  =  arc  PBP'  =  semicircle. 
Arc  PB     ='  arc  PP'  =  quadrant.  ^ 
Angle  POB  =  angle  BOP'  =  right 
angle. 

751.  Corollary.     On  a  sphere 
the  locus  of  a  moving  point  one  quad- 
rant distant  from  a  fixed  point  is  a  great  circle  having  the 
fixed  point  as  its  pole. 

Theorem  VIII. 

753.  The  poles  of  any  two  great  circles  lie  on  a 
third  great  circle,  having  their  points  of  intersection 
as  its  poles. 

Hypothesis.  AB,  CD,  two  great 
circles  intersecting  in  the  points  B  and 
P';  P,  P',  the  poles  of  AB;  Q,  Q',  the 
poles  of  CD. 

Conclusion.  The  poles  P,  Q,  P',  Q'  A 
lie  on  the  great  circle  having  E  and  P' 
as  its  poles. 

Proof.     1.  Because  P  and  P'  are 

■no^^+fl     '^^      ^lio     orroof.    nirnlo     A  Ti      of 

which  P  and  P'  are  the  poles. 

Arc  PR  =  arc  PB'=  arc  P'B  =  arc  P'P'=  quadrant  (§  749). 


THE  SPHERE. 


341 


2.  Because  R  and  R'  are  points  on  the  circle  CD  of 
which  Q  and  Q'  are  the  poles,  ' 

Arc  QR  =  arc  QR'  =  arc  Q'R  =  Q'R'  =  quadrant. 

3.  Because  P,  ^,  P',  and  Q'  are  points  each  one  quadrant 
distant  from  R  and  R',  they  lie  on  the  great  circle  having  R 
and  R*  for  its  poles  (§  761).    Q.E.D. 


B 


B 


Theorem  IX. 

•763.  Conversely,  if  the  poles  of  two  great  circles 
lie  on  a  third  great  circle,  the  two  great  circles  will 
intersect  in  the  poles  of  that  third  circle. 

Proof,    1.  Because  all  points  one  quadrant  distant  from  P 
lie  on  the  great  circle  AB,  the  poles  of 
every  great  circle  through  P  must  lie 
somewhere  on  the  circle  AB, 

2.  In  the  same  way,  the  poles  of 
eyery  great  circle  through  Q  lie  on 
the  great  circle  CD.  ^j 

3.  Therefore  the  poles  of  the  great 
circle  through  P  and  Q  lie  in  both  of 
the  great  circles  AB  and  CD)  that 
is,  in  the  points  R  and  R',  in  which 
AB  and  CD  intersect.     Q.E.D. 

•754.  Def  A  great  circle  having  a  point  B  as  its 
pole  IS  caUed  the  polar  circle  of  the  point  R. 

The  great  circle  containing  the  poles  of  other  great 
circles  IS  called  the  polar  circle  of  these  other  circles. 

^  ^5^,  Corollary  1.  If  any  number  of  great  circles  have 
tfieir  poles  upon  another  great  circle,  they  will  all  intersect 
tn  the  pole  of  that  other  circle. 

756.  Cor.  2.  If  any  numher  of  great  circles  pass  through 
a  common  point,  their  poles  will  all  lie  on  the  polar  circle  of 
that  -w^'****'  '^ 


842 


BOOK  X.    OF  CURVED  SURFACES. 


Theorem   X. 

757.  The  angular  distance  between  two  poles  of 
circles  is  equal  to  the  dihedral  angle  between  the 
planes  of  the  circles. 

Proof.  1.  If  P  and  Q  are  two  poles  of  circles,  then  the 
perpendiculars  from  P  and  Q  upon  p 

the  planes  of  the  circles  pass  through 
the  centre  of  the  sphere  (§§  730,  734).    c. 

2.  Because    these    perpendiculars 
pass  through  the  centre  of  the  sphere,  a[ 
the  arc  P^  is  equal  to  the  plane  angle 
between  them  (§  744). 

3.  Because  they  are  perpendicular 
to  the  planes  of  the  circles,  the  angle 
they  form  is  equal  to  the  dihedral  angle   between  those 
planes  (§  625). 

4.  Comparing  (2)  and  (3),  the  arc  P^  is  shown  to  be  equal 
to  the  dihedral  angle  between  the  planes.    Q.E.D. 

Theoeem    XI. 

758.  7/"  one  great  circle  passes  through  the  pole  of 
another,  their  planes  are  perpendicular  to  each  other. 

Proof.   1.  Because  one  pole  lies  on  the  polar  great  circle 
of  the  other  pole,  the  angular  dis- 
tance of  the  poles  is  a  quadrant. 

2.  Therefore  the  dihedral  angle 
between  the  planes  of  the  circles  is  a 
right  angle  (§  757).    Q.E.D. 

759.  Corollary.  If  any  number 
of  great  circles  vass  through  a  com- 
mon point,  the.,  planes  will  all  be 
perpendicular  to  the  plane  of  their 

"     -r-k  1         -1    rt    •!•  /o  iv(-j\     i^i     •        If  three  great  circles  Intersect 

For,     by    definition    (§  764),    their  at    Q,    their  planes    intersect 

polar  circle  passes  through  all  their  ^^ng  oq,  and  their  poles,  p, 

~    ,  -I    "i"       1  •      •i~       »  -^>  -^"'"5  lis  on   another  great 

poles,  and  its  plane  is  tneref'^re  per-  circle  of  which  q  is  the  poie, 
pendicular  to  all  of  their  planes  by  *'^<*  ^f  which  the  plane  opf'' 

*",        .  X  ./   jg  perpendicular  to  each  of  thf 

the  theorem.  given  planes. 


Illtistration  of  $$  758,  769. 


yoles  of 
3en  the 

■h&n.  the 


3n  those 
be  equal 


pole  of 
h  other. 
Bat  circle 


sles  Intersect 
*8  intersect 
lir  poles,  P, 
other  great 
is  the  pole, 
plane  OFF" 
each  of  thf 


THE  BPHERE. 


Theorem  XII. 


343 


•760.  Two  spherical  surfaces  intersect  each  other 
in  a  circle  whose  plane  is  perpendimlar  to  the  line 
joining  the  centres  of  the  spheres,  and  whose  centre 
is  in  that  line. 

Hypothesis.     0,  0',  the  centres  of  two  spheres;  ADB, 
the   curve    line    in  which 
their  surfaces  intersect. 

Conclusion.  ADB  is  a 
circle  having  its  centre  on 
the  line  00'  and  its  plane 
perpendicular  to  that  line. 
Proof.  Let  A  and  D 
be  any  two  points  on  the 
curve  of  intersection. 

Join  OA,  O'A,  OD,  and  O'D. 

From  A  and  D  drop  perpendiculars  upon  00\    Then— 
1.  In  the  triangles  OAO'  and  OBO'  the  side  00'  is 
common;  OA  =  OB  and  O'A  =  O'B,  because  these  lines  are 
radii  of  the  same  sphere.     Therefore 

Triangle  OA  0'  =  triangle  OBO'  identically. 
a.  Because  these  triangles  are  identically  equal,  perpen- 
diculars from  A  and  B  upon  the  base  00'  are  equal  (8  175) 
and  the  feet  of  these   perpendiculars  meet  00'  at  equal 
distances  from  0;  that  is,  at  the  same  point.     Let  C7be  this 
point. 

+  L?^T'^  *^^  .^'"^'  ^^  ^^^  ^^  ^^«  ^«*^  perpendicular 
/Lo.x  .1  ""^^  '"^  ''''^  P^^"®  perpendicular  to  this  line 
(^586);  and  because  they  are  equal,  their  ends  are  in  a  circle 
having  its  centre  at  C.     Q.E.D. 

Theorem  XIII. 
761.  A  plane  perpendicular  to  a  radius  of  the 


svhere  at  ita  ficrf.rfm'iUi  -/o  ^  ^ 


tyi^fui/c/oo  vu  c/ic  dpfiere. 


Proof.     Let  OP  be  the  radius  to  which  the  plane  is  per- 
pendicular.    Then— 


I 


'"  f- 


844 


BOOK  X.    OF  CURVED  SURFACES. 


1.  Because  OP  is  a  radius,  the  point  P  is  common  to  the 
sphere  and  the  plane. 

2.  Because  OP  is  perpen- 
dicular to  the  plane,  it  is  the 
shortest  line  from  0  to  the 
plane.  Therefore  every  other 
point  of  the  plane  is  without 
the  sphere. 

3.  Therefore  the  plane  is  a 
tangent  to  the  sphere  at  the  point  P.    Q.E.D, 

763.  Corollary  1.  Every  line  perpendicular  to  a  radius 
at  its  extremity  is  tangent  to  the  sphere  (§  725). 

763.  Cor.  2.  Conversely ,  every  plane  or  line  tangent  to 
the  sphere  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact.       \ 

Theorem  XIV. 

764.  All  lines  tangent  to  a  sphere  from  the  same 
external  point  are  equal,  and  touch  the  sphere  in  a 
circle  of  the  sphere. 

Proof.     Let  0  be  the  centre  of  the  sphere;  P,  the  point 
from  which  the  tangents  are  drawn; 
P8,  PT,  any  two  tangents  touching 
the  sphere  at  8  and  T.    Then — 

1.  In  the  triangles  PSO  and  P20 
the  side  PO  is  common  ;  OS  =  OT 
(because  they  are  radii),  and  angle 
OTP  =  angle  OSP  (both  being  right 
angles)  (§  763). 

Therefore  these  triangles  are  iden- 
tically equal,  whence 

PS  =  PT.     Q.E.D. 

2.  Because  the  triangles  PSO  and 
PTO  are  identically  equ2,  tlif  perpendiculars  from  8  xind  T 
upon  PO  are  equal,  and  meet  FO  at  the  same  dist8t{»30  from 
P\  that  is,  at  the  same  point  Q.  Since  8  and  T  may  bo  any 
two  points  in  which  tangents  through  P  touch  the  sphere,  all 
these  points  are  in  one  plane,  and  in  a  circle  having  its  coiifj-e 
at  Q.     Q.E.D 


/ 


THE  SPHEBB. 


Theorem  XV. 


.846 


765.    Through  four  points  not  in  the  same  plane 
one  spherical  surface,  and  no  more,  may  pass. 

Proof,  Let  ^^(7Z)  be  the  four  points. 
Join  ABy  BC,  CD,  and  bisect  each  of  these 
hnes  by  a  plane  perpendicular  to  them. 
Let  us  call  these  respective  planes  the 
planes  (ah),  {he),  {cd).    Then— 

1.  If  the  sphere  passing  through  A, 
B,  C,  and  D  exist,  then,  because  its  centre 
IS  equally  distant  from  A  and  B,  it  lies 
in  the  plane  {ah)  bisecting  AB  perpendicularly  (§  589). 

2.  In  the  same  way,  it  lies  in  the  other  two  bisecting 
planes,  and  therefore  in  their  point  of  intersection  if  they 
nave  one.  •' 

3.  If  they  have  no  point  of  intersection,  their  three  lines 
of  intersection  are  parallel  (§  637). 

4.  Suppose  these  lines  to  be  parallel.  Because  the  plane 
^jBC  contains  the  line  AB  1.  plane  {ah), 

Plane  ABC  L  plane  {ah),  (8  639) 

For  the  same  reason. 

Plane  ABC  ±  plane  {he). 
Because  plane  ABC  1  to  both  the  planes  (ah)  and  (he), 
and  (cd)  is  a  third  plane  having  parallel  lines  of  intersection 
with  (ah)  and  (he). 

Plane  ABC  1  plane  (cd).  (§  640) 

Because  plane  ABCL  plane  (cd),  and  line  CDJL  plane  (cd), 
by  construction,  therefore  CD  lies  in  the  plane  ABC(  §  631), 
and  A,  B,  C,  D  lie  in  one  plane,  which  is  contrary  to  the  hy- 
pothesis; whence  the  lines  of  intersection  are  not  parallel. 

5.  Therefore  the  three  planes  intersect  in  a  point  (§  637), 
which  point  is  equally  distant  from  A,  B,  C,  and  D,  and 
which  may  therefore  be  the  centre  of  a  sphere  passing  through 
A,B,  (7,  andi>.    Q.E.D.  ^        i'        6  s 

Corollarv.  The  fournoints  Ann  ot,/i  n^ — v«  a^v^- 
as  the  vertices  of  a  tetrahedron.  The  edges  will  then  be  the 
six  Unes  formed  by  joining  every  pair  of  vertices,  and  we  may 


'N: 


I 


iij 


346 


BOOK  X.     OF  CURVED  SUBFACES. 


take  any  three  of  those  edges  to  determine  the  positions  ol 
the  planes  whose  point  of  intersection  is  the  centre  of  the 
sphere.     Hence: 

766.  The  six  planes  which  bisect  at  right  angles  the  six 
edges  of  a  tetrahedron  all  pass  through  a  point* 


SIX 


Theorem   XVI. 

767.  ^  spJiere  may  be  tangent  to  any  four  planes 
wMcTi  do  not  intersect  in  a  point,  and  of  which  tJie 
lines  of  intersection  are  not  all  parallel. 

Proof.     Let  AB,  AC,  AD,  BG,  BD,  and  CD  be  the 
lines  of  intersection  of  the  four  planes, 
taken  two  and  two. 

Bisect  any  three  of  the  dihedral 
angles  which  lie  in  one  plane,  as  ABy 
BC,  and  AC,  by  other  planes. 

Let  0  be  the  point  in  which  these 
planes  meet.     Then — 

1.  Because  0  is  on  the  bisector  of 
the  dihedral  angle  BA,  it  is  equally 
distant  from  the  faces  ABC  and 
ABD  (§  638). 

2.  In  the  same  way  0  is  equally  distant  from  the  faces 
BCA  and  BCD,  and  from  CAB  and  CAD. 

3.  Therefore  the  point  0  is  equally  distant  from  the  four 
planes;  and  if  a  sphere  be  described  having  its  centre  at  0 
and  its  radius  equal  to  the  common  distant:,  it  will  be  tan- 
gent to  ail  four  planes.     Q.E.D. 

Corollary.  Since  we  may  take  any  three  dihedral  angles 
not  meeting  in  a  point  to  determine  the  centre  of  the  sphere, 
we  conclude: 

768.  The  six  planes  which  bisect  the  six  edge  angles  of  a 
tetrahedron  intersect  in  a  point. 

Scholium.  It  may  be  shown,  as  in  the  case  of  the  tri- 
angle, that  besides  the  sphere  inscribed  in  the  tetrahedron 
there  are  four  escribed  spheres,  each  touching  one  face 
externally  and  the  other  three  faces  produced. 


ions  of 
of  the 

the  six 


planes 
\ch  the 

the  six 


le  faces 

;he  four 
re  at  0 
be  tan- 

l  angles 
sphere, 

'?es  of  a 

the  tri- 
ahedron 
ne  face 


SPHERICAL  TRIANGLES  AND  POLYGONS  347 

CHAPTER    II. 

OF  SPHERICAL  TRIANGLES  AND  POLYGONS. 


769.  Def.  If  two  great  circles  of  the  sphere  inter- 
sect, they  are  said  to  make  an  angle  with  each  other 
equal  to  the  angle  between  their  tangents  at  the  point 
of  intersection. 

Illustration.     If  the  great  circles  a  and  I  intersect  at  0, 
and  if  H  and  K  are  their  respective  tan- 
gents at  Oy  then  their  angle  is  measured 
by  the  angle  HOK  between  the  tangents. 

770.  JDef.  A  spherical  triangle 

is  the  figure  fonned  by  joining  any 
three  points  on  the  sphere  by  arcs  of 
great  circles. 

771.  Def.  The  points  which  are 
joined  are  called  vertices  of  the  tri- 
angle. 

772.  Def.  The  arcs  of  which  a  spherical  triangle 
is  formed  are  called  sides  of  the  tri- 
angle. 


773.  Def  The  angles  which  the 
sides  make  with  each  other  are 
called  angles  of  the  triangle.  a  spherical  tnangie. 

Remark  1.  Between  any  two  points  two  arcs  of  a  gi-eat 
circle  may  be  drawn,  the  one  greater  and  the  other  less  than 
a  semicircle.  In  forming  a  spherical  triangle  the  arcs  less 
than  a  semicircle  are  supposed  to  be  taken,  unless  otherwise 
expressed. 

Remark  3.  In  a  spherical  triangle,  as  in  a  plane  tri- 
angle, each  side  has  an  opposite  angle  and  two  adianfiTi+. 
angles,  and  each  angle  has  an  opposite  side  and  two  adjacent 
Bides.  '' 


y."  aa}\ 


348 


BOOK  X.     OF  CURVED  SURFACES. 


Theobem  XVII. 


774.  The  angle  between  two  great  circles  is  equal 
to  the  dihedral  angle  betwa  i.ovoi  plants. 

Hypothesis.    AB,  CD,  iwo  g  cat  circles  of  the  sphere; 
PQ,  the  line  of  intersection  of  their 
planes;   QB,  QF,  their    respective 
tangents  at  Q. 

Conclusion.    The  angle  between 
the  circles  is  equal  to  the  dihedralAl 
angles  along  PQ. 

Proof.  1.  By  definition  the 
angle  between  thQ  circles  is  meas- 
ured by  the  angle  EQF  (§  769). 

2.  Because  QF  and  QE  are  tangents  to  the  circles  AB  and 
CD,  they  are  perpendicular  to  the  diameter  PQ;  and  because 
they  lie  in  the  planes  AB  and  DC,  their  angle  measures  the 
dihedral  angle  between  those  planes  (§  623). 

3.  Therefore  the  angle  between  the  great  circles  AB  and 
CD  is  equal  to  the  dihedral  angle  along  PQ.     Q.E.D. 

Theorem  XVIII. 

775.  J^  on  two  great  circles  points  he  taken  a 
quadrant  distant  from  their  points  of  intersection, 
the  arc  between  the  points  measures  the  angle  be- 
tween the  circles. 

Hypothesis.    AB,  CD,  two  great  circles  of  the  sphere 
intersecting  along  the  line  PQ',  B,  D, 
two  points  each  a  quadrant  distant 
from  P  and  Q. 

Conclusion.  The  arc  BD  is  equal 
to  the  angle  BQD  between  the  circles.  A(        '      >i^---l---^^ 

Proof.  From  the  centre  0  erect 
in  each  plane  a  perpendicular  to  PQ. 
Then— 

1.  Because    these    perpendiculars 
are  erected  from  the  centre,  they  will  meet  the  great  circle  in 
the  points  B  and  D. 


J  ^5  and 
L  because 
jures  the 

AB  and 
). 


taJcen  a 
^sectio% 
ngle  he- 

le  sphew 


III;  Circle  ii 


SPUmaOAL  TBI  ANGLES  AND  POLYGONS.  349 

2.  Because  OB  and  OD  are  radii  of  the  sphere, 

Arc  BD  =  angle  BOD. 

3.  Because  OB  and  OD  are  perpendicular  to  PQ^ 

Angle  BOD  =  dihedral  angle  of  planes. 

4.  Because  angle  BQD  =  dihedral  angle  of  planes  (§623), 
Angle  BOD  =  angle  BQD  =  angular  distance  BD.    Q.E.d! 

TnEOREM  XIX. 
•776.  ^^e  angular  distance  between  the  poles  of 
two  great  circles  is  equal  to  the  angle  between  the 
circles. 

Proof.  Let  P  aiid  ^  bo  the  poles 
of  the  great  circles  AB  and  OD,  and 
0  the  centre.   Join  OP,  OQ.  Then— 

1.  Because  P  and  Q  are  poles,  by  A 
definition. 


OP  1  plane  ^P; ) 


(§  734) 


Oq  1  plane  CD. 
Therefore 

Angle  PO^  =  dihedral  angle  between  planes, 

=  angle  between  circles.  Q.E.D. 

Corollary  1.  From  this  and  the  preceding  theorem,  with 
Theorem  XI.,  follows: 

7*77.  If  through  the  poles  P  and  Q  of  two  great  circles  a 
third  circle  be  passed  intersecting  the  other  circles  at  A,  0,  P, 
and  D,  we  shall  have — 

I.  Angle  PQ  =  angle  AC  =  angle  BD, 

=  angle  between  circles  AB  and  CD, 
=  dihedral  angle  BOD. 
II.  The  third  ci.  jle  will  intersect  both  the  other  circles  at 
right  angles. 

778.  Cor.  2.  If  two  sides  of  a  spherical  triangle  are 
quadrants,  the  angles  opposite  these  sides  will  be  right  angles. 

Belatioii  of  a  Spherical  Triangle  to  a  Trilie- 
dral  Angrle  at  the  Centre  of  the  Sphere. 

Because  the  three  sides  of  a  spherical  triangle  are  arcs  of 
51^0.0  unCiuo,  Luuir  pianea  ail  intersect  at  tne  eeniiu  of  the 
sphere,  where  they  form  a  solid  trihedral  angle. 


wm 


850 


BOOK  X.     OF  CURVED  SURFACES, 


By  §  744  tho  throo  sidoa  of  tho  triangle  are  measured  by 
the  three  plane  angles  of  tho  solid  angle,  and  by  §  774  the 
three  angles  of  the  triangle  are  measured  by  the  three  dihe- 
dral angles  of  tho  solid  angle.     Hence: 

779.  To  every  spherical  triangle  corresponds  a  trihedral 
angle  at  the  centre  of  the  sphere,  having  its  six  parts  equal  to 
the  parts  of  the  spherical  triangle. 

Conversely,  if  0-AliC  be  any  trihedral  angle,  we  may 
imagine  a  sphere  with  its  centre  at  0 
and  an  arbitrary  radius  OP,    Then — 

The  surface  of  tho  sphere  will  in- 
intersect  the  edges  OJ,  OB,  00  at 
the  points  P,  Q,  and  E. 

The  same   surface  will  intersect 
the  planes  OAB,  OIW,  OCA  in  tho 
arcs  of  great  circles  PQ,  QU,  HP, 
which  arcs  will  form  a  spherical  tri-  ^^ 
angle.    Hence: 

780.  Bvery  trihedral  angle  may 
be  represented  by  a  triangle  on  a 
sphere  having  its  ce?Ure  at  the  vertex. 

From  this  it  follows  that  the  relations  proved  in  §§  655-660 
between  the  faces  and  angles  of  a  trihedral  angle  are  true  of 
spherical  triangles,  when  for  the  face  angles  of  the  trihedral 
angle  we  substitute  the  sides  of  the  spherical  triangle,  and  for 
the  dihedral  angles  tho  angles  of  the  triangle.  We  thus  con- 
clude: 

'7 SI,  If  two  sides  of  a  spherical  triangle  are  equal,  the 
opposite  angles  are  equal  (§  655). 

782.  In  a  spherical  triangle  the  greater  side  and  the 
greater  angle  are  opposite  each  other  (§  656). 

783.  The  sum  of  any  two  sides  of  a  spherical  triangle  is 
greater  than  the  third  side  (§  659). 

784.  Two  spherical  triangles  are  equal  when  their  sides 
are  equal  and  similarly  arranged  (§  660). 

WOR       WT^    .^./v^.    «/•  47>a  iTiifoo    et'rfpa  r\fn  snlrp.r'Ir.nl,  f.rin.nnln 

is  less  than  a  circumference  (§  661). 


uurcd  by 
g  774  tho 
ireo  diho- 

trihedral 
8  equal  to 

we  may 


|§  655-660 
,re  true  of 
I  trihedral 
e,  and  for 
thus  con- 

equdly  the 

e  and  the 

riangle  is 

their  sides 


rtl  trianalo 


SPHBRICAL  TRIAmiKti  AND  POLYUOm.  351 

Symmetrical  Spherical  Triangles. 

786.  Def.  Two  spherical  triangles  are  said  to  be 
oppoaite  when  the  vertices  of  the  one  are  at  the  ends 
of  the  diameters  from  the  vertices  of  the  other. 

Corollary  1      Since  tho  radii  AO,  BO,  CO  from*  the  ver- 
tices  of  a  spherical  triangle  are  tho 
edges  of  the  trihedral  angle  correspond- 
ing to  it,  we  conclude: 

787.  To  two  oppoaite  spherical 
triangles  correspond  two  opposite  and 
symmetrical  trihedral  angles  (§  653). 

Corollary  2.     Since  tho  lines  AA\ 
BB\  and  CC  all  intersect  in  the  same 
point  0,  each  pair  of  them  is  in  one 
plane,  passing  the  centre  of  the  sphere     ^"^^^  ^^''  ^^''  ^^'  »»« 
and  containing  the  corresponding  arcs  Xtl^Lti'a^  S„Z"' 
AB  and  A'B',  5(7  and  B'C,  CA  and  CA\    Hence:       • 

788.  The  corresponding  sides  of  two  opposite  triangles 
are  formed  of  arcs  of  the  same  great  circle. 

^  789.  Symmetrical  spherical  triangles  are  those  in  which 
the  sides  and  angles  of  the  one  are  respectively  equal  to  those 
of  the  other,  but  arranged  in  the  reverse  order. 

Theorem  XX. 

790.  Opposite  spherical  triangles  are  symmetric 
cat. 

Proof.  Let  ABC  and  A'B'C  be 
the  triangles,  and  0  the  centre  of  the 
sphere.     Then— 

1.  Because  the  angles  AOB  and  •*■ 
A' OB'  are  in  one  plane,  we  have 

Angle  ^07?  =  opp.  angle  A' OB'-, 
and  for  the  same  reason, 
Angle  BOC  =  opp.  angle  B'OC; 
Anglo  GO  A  =  opp.  angle  C"OA'. 
3.  For  th«  same  reason. 


852 


BOOK  X.    OF  CURVED  SURFACES, 


Dihedral  angle  0^  =  dihedral  angle  0A\ 
Dihedral  angle  OB  =  dihedral  angle  OB', 
Dihedral  angle  0(7  =  dihedral  angle  0C\ 
Whence  the  corresponding  angles  of  the  two  triangles  are 
respectively  equal  (§  774). 

3.  To  an  eye  looking  from  0  the  yertices  A,  B,  C  and 
the  sides  AB,  BC,  CA  succeed  each  other  in  the  negative 
order  (§  648);  while  A',  B',  C  and  the  fides  A'B',  B'C, 
CA'  succeed  each  other  in  the  positive  order.  Therefore 
the  triangles  are  symmetrical.     Q.E.D. 

Note.  This  theorem  should  be  compared  with  §  653,  which  is  the 
equivalent  L'leorem  in  the  case  of'  a  trihedral  angle. 

791.  Corollary.  Two  symmetrical  triangles  cannot  in 
general  be  made  to  coincide  identically. 

For  if  we  slide  the  triangle  A'B'G'  over  to  the  other  side 
of  the  sphere  so  that  A' ~  A  and  C  =  C,  the  vertices  B'  and 
B  will  fall  on  opposite  sides  of  A  C. 

If  we  turn  one  triangle  round  so  that  B  and  B'  shall  fall 
on  the  same  side  of  AC,  the  vertex  A'  =  C and  C  =  A. 

Theorem  XXI. 
793.  If  two  symmetrical  spherical  triangles  are 
isosceles^  tJiey  are  identically  equal. 
Proof.   In  the  preceding  case  suppose 
Side  BA  =  side  BO; 
we  shall  then  have 

Angle  A  =  angle  0.  (§  781) 

Therefore,  in  the  symmetrical  triangle, 

Side  B'A'  =  side  B'C  =  side  BO  =  side  BA. 

Angle  A'  =  angle  0'    =  angle  0  =  angle  A. 

Then  if  we  slide  A'B'O'  over  so  that  ^'e  Cand  O'^A,  we 

shall  have 

Side  ^M'=  side  ^a 

^'ideB'G'  =  sideBA. 

Vertox  /?'  =  vertex  B. 

Therefore  the  two  triangles  are  identically  equal.    Q.E.D. 

Polar  Triangles. 
793.  Bef.  If  the  sides  of  one  spherical  triangle 
have  their  poles  at  the  vertices  of  a  second  triangle, 


r\ea  axe 

,  G  and 
legative 

bierefore 

ch  is  the 

nnot  in 

iher  side 
5  B'  and 

shaU  f aU 
A, 

lies  are 


(§781) 


I. 


E^,  we 


Q.E.D. 

triangle 
riangle, 


8PHERI0J  L  miANQLEa  Am)  POLYGONS.  353 

the  first  triangle  is  called  the  polar  triangle  of  the 
second. 

Theokem  XXII. 

794.  ThepoUr  triangle  of  a  polar  triangle  is  the 
original  triangle. 

Hypothesis.  ABC,  a  spherical  triangle;  A'B'C\  its  polar 
triangle.  , 

Conclusion.      The  polar    tri-  yK 

angle  of  A'B'O'  is  the  original  /'     ^'^^ 

triangle  ABC. 

Proof.  1.  Because  the  great 
circles  A'B'  and  A'C  have  their 
poles  at  G  and  B  (hyp.  and  def.),     ,         ,  ,        . 

their  point  of  intersection  A'  is   /        a' ^      \ 

the  pole  of  the  circle  BC  (§  753).  •^'' — ---"'"  ^ 

2.  In  the  same  way  it  is  shown  that  C"  is  the  pole  of  the 
circle  AB,  and  B'  ot  AC. 

3.  Because  the  three  great  circles  AB,  BC,  and  CA  have 
their  poles  at  C,  A',  and  i?',  ABCk,  by  definition,  the  polar 
triangle  of  ^'^'(7'.     Q.E.D.  ^ 

795.  Corollary.  The  relation  between  two  polar  tri- 
angles may  he  expressed  hy  saying  thai  the  vertices  of  each 
triangle  are  the  poles  of  the  sides  of  the  other. 

Theorem   XXIII. 

796.  In  two  polar  triangles  each  side  of  the  one 
is  the  supplement  of  the  opposite  angle  of  the  other. 

Hypothesis.  ABC,  A'B'C,  a  pair  of  polar  triangles  in 
which  A'B'  is  the  pole  of  the  c' 

vertex  C,  etc.  /^>kN 

Conclusions, 
Arc  AB    -\-  angle  C"  =  180°.  ^        X       \    Xfi* 

Arc  BC  -f  angle  /I'  =  180°. 
Arc  A'B'  -j-  angle  O  =  180°. 
Arc  B'C  -f  angle  A  =  180°.     / 

etc.        etc.        ttc.  jS[^>      -A^ 

Proof.     Produce  the  sides  « 

AB  and  A  C  until  they  meet  B' C  in  M  and  N.     Then— 


11 1. 

'it| 
1 11' 


!»}«=: 


364 


BOOK  X.    OF  CURVED  SURFACES. 


1.  Because  A  is  the  pole  of  MN,  AM  and  AN  are  quad- 
rants and  ,     ^  ._  ^^^. 

Arc  MN  =  angle  A.  (§  775) 

2.  Because  ^'  is  the  pole  of  ACN  and  C"  is  the  pole  of 

ABM, 

Arc  B'N=  arc  C*M—  quadrant,  or  90''. 

3.  Arc  B'C  -  arcs  i?'JV'4-  G'M  ~  JfiV  (identically); 
or  comparing  with  (2), 

Arc  B'C*-  180°  -  arc  MN\ 
and  comparing  with  (1), 

Arc  B'O'  -  180°  -  angle  A, 
4.  Therefore  arc  B'G'  +  angle  A  =  180°.    Q.E.D. 
In  the  same  way  all  the  other  relations  may  be  proved. 


? 


Theorem  XXIV. 

797.  The  sum  of  the  three  angles  of  a  spherical 
triangle  is  greater  than  a  straight  angle  and  less 
than  three  straight  angles. 

Proof.  Let  A,  B,  and  G  be  the  three  angles  of  the 
spherical  triangle,  and  a',  b',  and  c^  the  opposite  sides  of  the 
polar  triangle.     Then — 


I. 


A-\-a' 
B-\-h' 
G  +  c' 


=  130 
=  180 
=     180 


(§  796) 


Sum 
Whence 


A^B^G+a'  +  l'^-c'=  3.180° 
A  +  B^G=  3.180°  -  (a'  +  J'  +  c'). 
But  because  a',  J',  and  c'  are  sides  of  a  spherical  triangle, 
«'  _|_  J'  4-  c'  <  2.180°.  (§  785) 

Therefore  A  +  B  ■\-  G  >     180°.     Q.E.D. 

II.  Because  each  angle  of  the  triangle  is  less  than  a 
straight  angle,  the  sum  of  the  three  angles  is  less  than  three 
straight  angles.     Q.E.D. 

798.  Def.  The  spherical  excess  of  a  spherical  tri- 
angle is  the  excess  of  tlie  sum  of  its  three  angles  over 
a  straight  angle. 

Gorollary.  The  spherical  excess  may  de  equal  to  any 
positive  angle  less  than  a  circumference. 


f 


Tm  OTLINDER 


355 


e  quad- 

(§  'S'75) 
pole  of 


CHAPTER  ill. 

THE  CYLINDER. 


ically); 


ved. 


Tierical 
%d  less 

of  the 

38  of  the 


(§  ^96) 


briangle, 
(§  ^85) 

than  a 
an  three 

ical  tri- 
les  over 

to  any 


799.  Def.  A  cylindrical  surface  is  the  surface 
which  is  generated  by  the  motion  of  a  straight  line 
constantly  touching  a  given  curve,  and  remaining 
parallel  to  its  original  position. 

Illustration.    If  the  straight  line  AB  move  around  the 
curve  AM,  remaining  parallel  to  the  posi- 
tion AB  during  the  motion,  it  will  generate  ^ 
a  cylindrical  surface. 

800.  Def.    The  generatrix  is  the 
line  which  generates  the  surface. 

801.  Def.    The  directrix  is  the 
curve  which  the  generatrix  touches.      ^ 

802.  Def.  A  cylinder  is  a  solid 
bounded  by  a  cylindrical   surface   and  two  parallel 
planes, 

803.  Def.  Elements  of  the  cylinder  are  the  differ- 
ent positions  of  the  generatrix. 

Remaek  1.  In  elementary  geometry  the  directrix  is  a 
circle  whose  plane  is  perpendicular  to  the  generatrix. 

Kemark  3.  Since  the  generatrix  may  extend  out  indefi- 
nitely in  two  directions,  a  cylindrical  surface  may  extend  out 
indefinitely  in  two  directions. 

804 .  Def.  The  axis  of  a  cylinder  is  a  line  through 
the  centre  of  the  directrix  parallel  to  the  elements. 

805.  Def.  A  tangent  plane  to  a  cjdinder  is  one 
which  touches  the  cylindrical  surface  without  inter- 
secting it. 

806.  Def.  xV  right  section  of  a  cylinder  is  the  sec- 
tion by  a  plane  perpendicular  to  the  elements. 

807.  Def.  A  sphere  is  said  to  be  inscribed  in  a 
cylinder  when  all  the  elements  of  the  cylinder  are 
tangents  to  the  sphere. 


■I' 


366 


BOOKX.    OF  CURVED  SURFACES. 


Theoeem  XXV. 

808.  A  plane  tangent  to  a  cylinder  is  parallel  to 
all  the  positions  of  the  generatrix,  and  touches  the 
cylindrical  surface  along  an  element. 

Hypothesis.    KLMN^  a  cylindrical  surface;  P,0',  two 
points  in  which  a  plane  may  touch  the 
surface  without  intersecting  it. 

Conclusion.  The  line  P^'  is  an  ele- 
ment and  lies  in  the  plane,  and  all  other 
elements  are  parallel  to  the  plane. 

Proof.   1.  Let  KM  be  any  element.  If 
KM  were  not    parallel    to  the  trngent 
plane,  it  could  be  produced  so  far  as  to  Mt^ 
intersect  the  plane,  and  then  the  cylindri- 
cal surface  would  also  intersect  the  plane,  which  is  contrary 
to  the  definition  of  a  tangent  plane.     Therefore 
KM  II  tangent  plane.     Q.E.D. 

2.  Let  P^  bo  the  element  which  passes  through  P.  Be- 
cause P  is  in  the  tangent  plane,  and  PQ  cannot  intersect  the 
plane  (1),  Q  is  in  the  plane.  But  Q'  is  also  in  the  plane,  by 
hypothesis.  Then  the  plane  would  intersect  the  curve  MN 
at  the  points  Q  and  Q',  and  therefore  would  intersect  the 
surface  also,  which  is  contrary  to  the  definition  of  a  tangent 
plane. 

3.  Therefore  the  points  Q  and  Q'  are  the  same,  and  PQ' 
is  an  element. 

4.  Because  P  and  Q  are  both  in  the  plane,  the  line  P^  is 
in  this  plane,  and  is  the  line  of  contact  between  the  cylinder 
and  plane.    Q.E.D. 


Theoeem  XXVI. 

809.  If  a  sphere  he  inscribed  in  a  cylinder,  the 
^irfaces  will  touch  on  a  great  circle  the  plane  of 
which  forms  a  right  section  of  the  cylinder. 

Proof.  Let  0  be  the  centre  of  ttne  sphere,  and  A3  any 
element  of  the  cylinder.     Through  0  pass  a  plane  perpen- 


f  t 


THE  CYLINDER, 


857 


frallel  to 
cTies  the 


I  P.    Be- 

ersect  the 
plane,  by 
irve  MN 
irsect  the 
\  tangent 

and  PQ' 

ne  PQ  is 
)  cylinder 


ider^  tlie 
')lane  of 

I  AB  any 
s  perp€n- 


diculM-  to  the  elements  of  the  cylinder,  which  call  the  plane 
(/.     Then —  ^ 

1.  Because  the  sphere  is  inscribed  in  the 
cylinder,  the  line  AB  is  tangent  to  the  sphere 
(§  807).  ^ 

2.  Let  P  be  the  point  of  tanffencv.  Join 
OP,    Then  ^ 

OP  L  AB.  (§  763) 

Therefore  AB  being  X  plane  0,  OP  lies  in 
the  plane  0;  whence  P  also  lies  in  this  plane, 
and  is  the  only  point  in  which  AB  meets  the 
surface  of  the  sphere. 

+n  \  .^""^T  ^^  T^  ^'  ^""^  ^^'^^^*'  a"  tlie  elements 
touch  the  sphere  m  the  plane  0  perpendicular  to  them 

4.  Because  the  plane  0  passes  through  the  centre  of  the 
sphere  and  the  points  of  tangency  are  all  on  its  intersection 
with  the  surface,  these  points  are  on  a  great  circle  of  the 
spnere.     Q.lli.D. 

o.V^7fZ^'  It  *?  'P^''"'  ^''  ms^nh^^di  in  the  same 
cylmder,  then  the  planes  of  contact,  being  perpendicular  to 

spheres  are  m  the  axis  perpendicular  to  the  planes,  their  dis- 
tance IS  also  equal  to  the  distance  between  those  centres 
Hence: 

lenffl^n'f  fr'^^^'f  '""''f'^  *^  i^^^  ^^rne  cylinder  intercept 
lengths  of  the  elements  equal  to  the  distance  letioecn  the  centres 
of  the  spheres.  ^'"i^/vd 

Theorem  XXVII. 
i.  n^l)'^'"''^ ^^""''^  ''^^'"^  ^^^  <^y^^'dTical surface 

i^J!^^'^^''''''     ^^^''  '"'^  ^^^'''  '''*^^''  ""^  *  cylindrical  sur- 

Conclusion.     APB  is  an  ellipse. 

Proof.  In  the  cylinder  inscribe  two  spheres  of  which 
^  and  Q  are  the  centres,  in  such  a  position  that  each  of  them 
snail  touch  the  piano  A .« 

Let  ^-and  F  be  the  points  of  contact  with  the  plane. 
LBt  P  oe  any  point  oi.  the  section  APB,  HI  the  element 


'M 


358 


BOOK  X.    OF  CURVED  8USFACE8. 


passing  through  P,  and  H  and  I  the  points  at  which  this 
element  touches  the  spheres  0  and  Q. 
Join  PF  and  PB.    Then— 

1.  Because  the  plane  ^^  is  tangent  to 
the  sphere  0  at  B,  and  P  is  a  point  in  this 
plane,  PB  is  a  line  tangent  to  the  sphere 

at^. 

2.  Because  P/f  is  another  tangent  from 
P  to  the  same  sphere, 

PB  r=.  PH.  (§  764) 

3.  We  find  in  the  same  way,  for  the  A 

sphere  Q, 

PF-PL 

Whence 

PE-\-PFr:z  PII^  PI=  HI, 

or  PF  +  PF=OQ,  (§810) 

4.  Since  P  may  be  any  point  of  the  section,  the  sum  of 
the  distances  of  every  point  of  the  section  from  F  and  F  is 
equal  to  the  same  constant  length  OQ.  Therefore  the  sec- 
tion is,  by  definition,  an  ellipse  around  ^and  Pas  foci  (§  614). 

^  Q.E.D. 

813.  Corollary  1.  The  distance  of  the  centres  of  the  in- 
scribed tangent  spheres  is  equal  to  the  major  axis  of  the  ellipse. 

813.  Cor.  2.  The  tangent  spheres  touch  the  plane  of  the 
ellipse  at  its  respective  foci. 

814.  Cor.  3.  Parallel  plane  sections  of  a  cylindrical 
surface  are  identically  equal. 


»  » 


CHAPTER    IV. 

THE  CONE. 


815.  Bef.  A  conical  surface  is  the  surface  gener- 
ated by  the  motion  of  a  straight  line  which  constantly 
passes  through  a  fixed  point  and  touches  a  curve. 


TEE  CONE. 


359 


ich  this 


h 


B 


Hi 


e  snm  of 
ind  F  is 
the  sec- 
i  (§  614). 
i.E.D. 
f  the  in- 
le  ellipse. 

me  of  the 
lindrical 


se  gener- 
nstautiy 
rve. 


816.  A  cone  is  the  solid  formed  by  cutting  off  a 
portion  of  a  conical  surface  by  a  plane. 

A  cone  is  completely  bounded  by  the  conical  surface  and 
the  plane. 

817.  The  base  of  a  cone  is  its  plane  surface. 
Remark.    In  the  higher  geometry  a  conical  surface  is 

called  a  cone,  and  we  shall  use  this  abbreviation  when  con- 
venient. 

818.  Def.  The  generatrix  is  the  generating  line 
of  a  cone. 

819.  JDef.  The  directrix  of  a  cone 
is  the  curve  along  which  the  generatrix 
moves. 

830.  The  vertex  of  a  cone  is  the 
fixed  point  through  which  the  genera- 
trix passes. 

831.  Bef.    Elements  of  the  cone        a  cone, 
are  the  straight  lines  occupying  the  different  positions 
of  the  generatrix. 

Remark  1.  In  elementary  geometry  the  directrix  is  sup- 
posed to  be  a  circle. 

Rema  iK  2.  Since  the  generating  line  may  extend  on  both 
sides  of  tne  vertex,  a  complete  conical  surface  consists  of  two 
surfaces  meeting  in  a  point  at  the  vertex  and  extending  out 
indefinitely  in  both  directions. 

833.  Def.  The  two  parts  of  a  complete  conical 
Suixace  are  called  nappes  of  the  cone. 

833.  Def.  A  tangent  plane  to  a  cone  is  a  plane 
touching  the  conical  surface  without  intersecting  it. 

834.  Def    A  sphere  is  said  to  be  inscribed  in  a 

cone  when  ail  the  elements  of  the  cone  are  tangents  to 
the  sphere* 


iLviii 


m 


^'m 


'iVi 


360 


BOOK  X.    OF  CURVED  aUBFACES. 


if 


I  i 


11 


825.  D^.  The  axis  of  a  cone  is  the  straight  line 
from  the  vertex  through  the  centre 
of  the  directrix. 

836.  Def.  A  right  cone  is  one 

in  which  the  vertex  is  in  the  line 
passing  through  the  centre  of  the 
directrix  perpendicular  to  its  plane. 

Note.  In  the  following  propositions  the 
word  cone  means  a  right  cone,  though  eome  of 
the  theorems  are  true  of  other  cones. 

Theorem  XXVIII. 

827.  A  plane  tangent  to  a  cone  touches  it  along 
an  element,  and  passes  through  the  vertex. 

Hypothesis.  0-AB,  a  conical  surface;  M,  a  point  at  which 
a  plane  touches  the  surface. 

Conclusions. 
I.  The  plane  passes  through  0, 

II.   OMj  and  no  otlier  element,  lies  in  it. 

Proof.  I.  If  the  plane  did  not  pass 
through  0,  then,  because  it  does  pass 
through  M,  OM  would  intersect  the  plane  at 
My  and  the  plane  could  not  be  a  tangent. 

Therefore  the  plane  passes  through  0. 

Q.E.D. 

II.  Because  the  points  0  and  M  both  lie  in  the  plane,  the 
element  OM  lies  wholly  in  it. 

If  any  other  element  than  OM  may  lie  in  the  plane,  let 
OA  be  that  element.  The  plane  would  meet  the  directrix 
AB  vX  two  points  A  and  K,  and  therefore  would  intersect  it 
and  would  not  be  a  tangent  plane. 

Therefore  the  plane  touches  the  cone  only  along  the 
element  OK,     Q.E.D. 

Scholium,  In  II.  of  this  demonstration  it  is  assumed  that 
no  part  of  the  directrix  is  a  straight  line.  If  such  were  the 
case,  a  portion  of  the  conical  surface  would  be  a  plane,  and 
the  tangent  plane  might  coincide  with  this  plane  surffice. 


' 


lit  line 


t  along 
kt  which 


ano,  the 

lane,  let 
iirectrix 
ersect  it 

ong  the 

aed  that 
wrere  the 
a.ne,  and 
fice. 


THE  GONE. 


Theorem  XXIX. 


361 


828.  If  a  sphere  he  inscribed  in  a  cone,  the  sur^ 
faces  touch  on  a  circle  all  points  of  which  are  equally 
distant  from  tJie  vertex. 

^     Proof.  When  a  sphere  is  inscribed  in  a  cone,  each  element 
IS,  by  definition,  a  tangent  to  the  sphere. 
Hence  all    the    elements   are   tangents 
to  the  sphere  from  the  vertex,  and  are 
equal  by  Theorem  XV.  (§  764). 

From  the  same  theorem  it  follows 
+hat  the  points  of  contact  lie  upon  a 
circle.     Q.E.D. 

839.  Corollary.  If  hoo  spheres  be 
inscribed  in  the  same  cone,  the  segments 
of  the  elements  intercepted  between  the 
points  of  tangency  are  all  equal. 


Theorem  XXX. 

830.  Bmry  complete  plane  section  of  one 
of  a  cone  is  an  ellipse. 

Hypothesis.  APB,  a  plane  section  of  one  nappe  of 
passing  through  all  the  elements. 

Conclusion.   APB  is  an  ellipse. 

Proof  Let  0  and  Q  be  the  centres 
of  two  spheres  mseribed  in  the  cone, 
and  tangents  to  the  cutting  plane  at 
the  respective  points  E  and  F. 

Let  P  be  any  point  of  the  section, 
and  a  and  H  the  points  in  which  the 
element  VP  touches  the  respective 
spheres. 

Join  PE,  PF.     Then-. 

1.  Because  PE  and  PG  are  tan- 
gents to  the  aume  sphere, 

PE  =  PG, 


nappe 


86S 


BOOK  X.    OF  VUBYED  SUBFACES, 


2.  In  the  same  way, 

whencfc 

PF\-PF=OH. 

3.  Because  0  and  H  are  the  points  in  which  an  element 
touches  two  inscribed  spheres,  the  line  OH  has  the  same 
length  for  all  the  elements  (§  829). 

Therefore  the  sum  of  the  distances  PB  -f  PF  is  the  same 
for  every  point  of  the  section,  and  this  section  is  an  ellipse,  ])y 
definition.     Q.E.D. 

831.  Corollary.  The  points  in  which  the  inscribed 
spheres  touch  the  cutting  plane  are  the  foci  of  the  ellipse. 

Theokem  XXXI. 

832.  E:i}ery  section  of  hotli  nappes  of  a  cone  by  a 
plane  is  an  hyperbola. 

Hifpothesis.  AB,  a  section  of  two  nappes  of  a  cone  by  the 
same  plane;  F,the  vertex  of  the  cone. 

Conclusion.  AB  is  an  hyper- 
bola. 

Proof.  Let  0  and  Q  be  the  cen- 
tres of  two  spheres  inscribed  in  the 
cone,  and  tangents  to  the  cutting 
plane  at  the  points  F  and  F. 

Let  P  be  any  point  of  the  sec- 
tion; PVG,  the  element  through 
P;  and  G  and  H,  the  points  in  which 
this  element  touches  the  spheres. 

Join  PF,  PF    Then— 

1.  Because  the  plane  AB  is  tan- 
gent to  both  spheres  at  the  points 
F  and  F,  and  P  is  in  this  plane, 

PF  and  PF  are  tangents  to  the 
respective  spheres. 

2.  Because  PF  and  PG  are  tangents  to  the  sphere  0  from 
the  same  point  P,  PF  =  PG. 


rt  in 


and 


r>  TT 


-«  i-^. 


»j%»»-4-rt  4-^^   ■i'T% «^ 


nif\  V%  ^\*ty\ 


n 


o.  Decause  1''jc'  Hud.  I'ti  arc  langenis  lo  me  spnero  \^y 


PF  =  PH. 


THE  CONE. 


803 


4.  Subtracting  this  equation  from  (2), 

PE--PF=^  PG-PH=  0H=  0V-{-  VH, 

6.  But  the  lengths  VQ  and  VH  are  each  constant  for  all 
the  elements  of  the  cone.  Therefore  since  every  point  of  the 
section  must  bo  on  some  element  of  the  cone,  the  difference 
of  the  distances  of  all  such  points  from  i^and  Fib  the  same. 

fj.  Therefore  the  section  is  an  hyperbola  Laving  its  foci  at 
^andi^(§531).     ^  E.D. 

833.  Corollary.  TJie  major  axis  of  the  hyperMa  is 
equal  to  the  common  length  of  the  segments  of  the  elements 
contained  between  the  points  in  which  they  touch  the  inscribed 
spheres. 

Scholium.  Let  MNR8  be  a  portion 
of  a  conical  surface,  of  which  AB'iq  the 
axis,  extending  out  indefinitely;  andM< 
let  an  indefinite  plane  pass  through  a 
point  X     Then— 

1.  If  the  plane  makes  with  the  axis 
an  angle  greater  than  A  VH,  it  will  in- 
tersect one  nappe  of  the  cone  com- 
pletely, and  will  not  touch  the  other 
nappe.    The  section  is  then  an  ellipse. 

2,  If  the  plane  makes  with  the  axis 
an  angle  less  than  that  of  the  cone,  it 
will  partly  intersect  both  nappes  but 
will  cut  through  neither  of  them. 
The  section  is  then  an  hyperbola. 

?•  ?i^®  ?^T  ""^^^^  ^'^^  ^^®  ^^^s  an  angle  equal  to  the 
angle  AVM  of  the  cone,  it  will  cut  into  one  nappe  of  the  cone 
indefinitely,  but  will  not  cut  the  other.  If  we  imagine  the 
plane  ^Xto  turn  upon  X  until  it  assumes  the  position  PX, 
the  lower  of  the  two  tangent  inscribed  spheres  will  be  moved 
off  indefinitely.  Hence  the  focus  of  the  elliptic  section  of 
the  cone  will  move  off  indefinitely,  and  the  ellipse  will  become 
a  parabola  (§  560).     Hence: 

834.  Corollary.  The  section  of  a  cone  bv  a  vlane  mralM 
io  an  element  is  a  parabola,  "     " 


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BOOK  XL 
THE  MEASUREMENT  OF  SOLIDS. 


CHAPTER    I. 

SUPERFICIAL    MEASUREMENT. 


835.  Def.  A  right  parallelepiped  is  a  parallelo- 
piped  in  which  the  lateral  faces  are  at  right  angles  to 
the  base.         \ 

Remark  1.  Since  any  face  may  be  taken  as  the  base,  any 
parallelopiped  in  which  one  face  is  at  right  angles  to  the 
four  adjoining  faces  is  a  right  parallelopiped. 

Kemark  2.  A  right  parallelopiped  differs  from  a  rect- 
angular one  (§  685)  in  that  its  base  need  not  be  a  rectangle. 

836.  Bef.  The  lateral  area  of  a  prism  or  pyramid 
is  the  combined  area  of  its  lateral  faces  (§§  672,  695). 

837.  Def,  A  prism  is  said  to  be  inscribed  in  a 
cylinder  when  its  bases  coincide  with  the  bases  of  the 
cylinder,  and  its  lateral  edges  are  elements  of  the 
cylinder. 

838.  Def.  A  pyramid  is  said  to  be  inscribed  in  a 
cone  when  its  base  is  a  polygon  inscribed  in  the  base 
of  the  cone,  and  its  lateral  edges  are  elements  of  the 
cone. 

839.  Def.  A  regular  pyramid  is  a  pyramid  of 
which  the  base  is  a  regular  polygon,  the  perpendicu- 
lar through  whose  centre  passes  through  the  vertex 
of  the  Dvramid. 


SUPERFICIAL  MEA8UBEMENT. 


366 

Theorem  I. 
841.  The  lateral  area  of  a  prism  t<i  Pm,ni  /^  *% 

Proof,  Let  ABUS  he  any  lat. 
eral  face  of  a  prism,  and  KLMNP 
a  right  section  (§  674).     Then— 

1.  Because  the  lines  LM,  MN 
NP,  etc.,  which  maJce  up  the  peri' 
meter  of  the  section  are  all  in  one 
plane  perpendicular  to  the  parallel 
edges  AR,  B8,  etc.,  they  are  per- 
pendicular  to  these  edges. 

2.  Because  ABR8  is  a  paral- 
lelogram of  which  XJf  may  be  taken  a^  an  altitude, 

^Tea,ABES=LM.BS.  ^8  295^ 

takin^^hr''  ^t.!"'''''^  '^^''  ^^^  ^"  '^^^^'  ^e  find,  by 
taking  the  sum  of  those  areas  formed  in  the  same  way. 
Lateral  surface  =  (KL  +  XJf  +  MJV-\-  NP  +  pi^^'. 

0  E  D 
^"^o^ry     Because  the  base  of  a  right  prism  is  a  right 
section,  and  Its  altitude  a  lateral  edge:  ^ 

muu^'iJ^tj,  ^'"""^  r""  "■^''  "^'''  ^"^  ««  «?««^  '0  its 

mitms  into  tJie  perimeter  of  its  base. 

Theoeem  n. 
fo^^fL5f!.f«  "■^'^  oyUndrical  miface  is  equal 

section  of  this  prism.    Then-'  """  '^"  ^^^"^^^  "«  ^  "g^^t 


;1 


366 


BOOK  XL    MEASUREMENT  OF  80LID8. 


1.  Because  the  prism  is  inscribed  in  the  cylinder,  each 
lateral  edge  will  be  equal  in  length  to  an 
element  of  the  cylinder,  and  will  lie  in 
the  cylindrical  surface  (§  837). 

2.  Therefore  the  lateral  surface  of 
the  prism  will  be  equal  to  the  perimeter 
of  the  cross-section  AD  into  the  length 
of  each  element. 

Now  suppose  the  number  of  sides  of 
the  prism  to  be  increased  without  limit. 
Then— 

3.  The  x^rhneter  of  the  cross-section 

AD  (which  call  P)  will  approach  the  circumference  of  the 
cross-section  of  the  cylinder  (which  call  0)  as  its  limit  (§482). 

4.  The  lateral  surface  of  the  prism  (which  call  S)  will 
approach  the  surface  of  the  cylinder  (which  call  6")  as  its 
limit. 

6.  Because  we  continually  have  S  =  P  X  length  of  ele- 
ment, however  great  the  number  n  of  sidas,  and  because  P 
approaches  C  as  its  limit  and  S  approaches  /S"  as  its  limit,  we 
have  S'  =  G  X  length  of  element.  Q.E.I>. 


Theoeem  III. 

844.  The  lateral  area  of  a  regular  pyramid  is 
equal  to  half  its  slant  height  into  the  perimeter  of  its 
base. 

Proof.  Let  V-ABCDE  be  a  regular  pyramid,  and  VN 
its  slant  height.    Then— 

1.  Considering  ^J5  as  the  base  of 
the  triangle  VAB,  FJVwill  be  its  alti- 
tude.   Therefore 

Area  VAB  =^  ^VK ,  AB. 

2.  Because  the  slant  heights  are  all 
equal  to  FJV", 

Area  VBC  =  iFJV^.  BC, 
Area  VCD  =  iFiV^.  CD, 


etc. 


etc. 


SUPERFICIAL  MEASURBMEMT. 


367 


3.  Taking  the  sum  of  all  these  areas, 

Lateral  area  =  i  VN{AB  +  ^C+  CD  +  etc.), 

-  iFiV.  perimeter  ABODE.    Q.E.D. 

Theoeem  IV. 
846.  The  lateral  area  of  a  frustvm  ni'  n  ^^....i 
pyramid  is  equal  to  its  slant  heigkt  ^^      ^''^'' 

trUo  half  the  sum  of  the  perimeters       F^-i^i 
of  its  bases. 

Proof,   Let  ABCDH-FOmj  he  the 
frustum.    Then— 

1.  Because  the  lateral  faces  are  trape-  ^ 
zoids,  having  FG  ||  AB,  OH  \\  BO,  etc.,      b C 

8.  Taking  the  sum  of'all  these  1^^,  ^^  '**^> 

Lateral  area  =  *  (AB  +  BC+  etc.  ^  FG  +  GH+  etc.) 

X  slant  height.  ' 

:^«  t  f  21 1 1"-  =  ?«"'»«'«■•  of  lower  base; 
Therefore     "'  ""  perimeter  of  upper  base. 

Lateral  area  =  i  sum  of  perimeters  x  slant  height.    Q.B.D. 

Theobem  V.         * 

n-  „  ^     -^  .,,     .   ^J^^ifJ^J^  be  the  right  cone.     Inscribfi  in 
It  a  pyramid  having  a  regular  base  of  any 
number  of  sides.     Then— 

1.  Because  the  edges  of  the  pyramid 
coincide  with  the  elements  of  the  cone 
they  are  all  equal  and   the    pyramid    is 
regular.     Therefore 
Lateral  area  of  pyramid  =  ^  slant  height 

^  X  perimeter  of  base. 

,     -^et  the  number  of  sides  be  indefinitely 
increased.     Then—  ^ 


868 


BOOK  XL    MEA8  UREMENT  OF  SOLIDS, 


2.  The  perimeter  of  the  base  of  the  pyramid  will  approach 
the  circumference  of  the  base  of  the  cone  as  its  limit. 

3.  Therefore  th')  slant  height  of  the  pyramid  will  approach 
the  slant  height  of  the  cone  as  its  limit. 

4.  The  surface  of  the  pyramid  will  approach  the  surface 
of  the  cone  au  its  limit.    Therefore 

Lateral  area  of  cone  =  i  slant  height  X  circumf.  of  base. 

Q.E.D. 

847.  Corollary  1.  The  lateral  area  of  the  frustum  of  a 
right  cone  is  equal  to  the  product  of  its  slant  height  into  half 
the  sum  of  the  circumference  of  its  bases. 

Cor.  2.  If  the  frustrum  of  the  pyramid  be  cut  midway 
by  a  plane  parallel  to  the  base,  the  section  of  each  face  will 
be  half  the  sum  of  the  bottom  and  top  edges  (§  170).  Hence 
the  perimeter  of  the  section  will  be  half  the  sum  of  the  perime- 
ter of  its  bases'.   The  same  will  be  true  of  the  cone.  Hence: 

848.  The  lateral  area  of  a  frustum  of  a  cone  is  equal 
to  its  slant  height  into  the  circumference  of  its  mid-section. 

Spherical  Areas. 

Theorem  VI. 

849.  Two   symmetrical  spherical  triangles  are 

equal  in  area.  * 

Proof.    Let  ABC  and  A'B'C  be  the  two  symmetrical 

triangles  placed  opposite  each 
other  (§786);  P,  the  pole  of  the 
circle  of  the  sphere  passing 
through  A,  B,  and  C.  From  P 
draw     the    diameter     POP'. 

Then— 

1.  Because  P  is  the  pole  of 
the  circle  through  A,  B,  and  G, 

Arc  PA  -  arc  PB  =  arc  PC. 

(§  ^45) 

2.  Because  A',  P',  C\  and 
P'  are  opposite  points  to^,  B,C,  and  P,  respectively, 

P'^'  JPA,    P'B'  =  PB,^  P*C'  =  PC', 
whence  P'A'  =  P'B'  =  r'C\ 


SUPERFICIAL  MEASUREMENT. 
3.  Therefore  ^PCand  A'P'C'  BPA  and  B'P'A  ' 


369 


/nr 


Ta  -a    ?\T  ^^^P^c^ively  isosceles  symmetrical  triangles 
and  identically  equal  (§§  784,  792).  ^ 

4.  Because  the  sum  of  the  three  triangles  APO    etc 
makes  up^^C;  and  the  sum  of  the  three  equal  trianglj^ 
^  P'5',  etc.,  makes  up  ^'^'C,  therefore 

Area  A'B'C  =  area  ABC.    Q.E.D. 

850.  Def  A  lune  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  two  great  semi- 
circles. 

^  861.  Be/.  The  angle  of  the  lune 
IS  the  angle  between  the  great  semi- 
circles which  bound  it. 

Corollary.  The  angle  of  a  lune  is  equal 
to  the  dihedral  angles  between  the  planes 
of  its  bounding  semicircles  (§  774).  Ahae, 

Theorem  VII. 
852.  On  the  same  sphere,  or  on  equal  spheres 
lunes  of  equal  angles  are  identically  equal. 

Proof.  Let  the  two  spheres  be  applied  so  that  their  centres 

Turn  one  sphere  ronnd  on  its  centre  so  that  the  vertex 
and  one  semicircle  ol  itelune  shall  coincide  with  the  lo^ 
spending  vertex  and  semicircle  of  the  other 

Because  the  angles  are  identically  equal,  the  planes  of  the 
two  semicirc  es  will  then  coincide,  and  therefore  the  bound! 
mg  semicircles  will  also  coincide.  The  lunes  4  therefore 
identicaUy  equal,  by  definition.     Q.E.D.  werelore 

Theorem  VIII. 

n^^^'  '^^J^^'f^  O^^t  '^^eiee  mutually  intersect 
the  areas  of  the  two  triangles  on  opposite  sides  of 
rniy  vertex  are  together  eq,ml  to  the  area  of  a  lune 
•i..,..^vy  u,v^  ^ivyvvjvi iiita  ai  me  vertex. 


■SH 


370 


BOOK  XL    MEASUUBMENT  OF  SOLIDS. 


Hypothesis.    AB,   Gil,  MN,  any  throo  great  circles;  P, 
any  vortex  where  two  of  them  crosa. 

Conclusion. 
Areas  FGM+FIIN=  ImioFHQNP. 

Proof.  1.  Because  PMG  and  QNN 
are  opposite  triangles  formed  between  gI 
the  same  great  circles,  they  are  sym- 
metrical triangles  (§  790). 

Therefore 

Area  PMG  =  area  QNH. 

2.  Therefore 
Area  Pmi  +  area  PMG  =  area  PNH-\-  HNQ, 

=  area  of  lune  Pi/(?iVP.    Q.E.D. 

Theorem  IX. 

854.  The  area  of  a  lune  is  to  the  whole  surface 
of  the  sphere  as  the  angle  of  the  lune  to  a  circumfer- 
ence. 

Proof.  1.  Let  PAQB  be  a  lune  of  which  P  and  Q  are  the 

vertices. 

From  the  vertex  P  to  ^  pass  n  semi- 
circles making  equal  angles  with  each 
other.  Th'^  whole  surface  of  the  sphere 
will  then  be  divided  into  n  equal  lunes. 

(§  853) 

2.  Let  m  be  the  number  of  those 
equal  lunes  contained  in  the  given  lune 
PAQB.  The  ratio  of  this  lune  to  the 
surface  of  the  sphere  will  then  be  w  :  ». 

3.  The  angle  APB  is  made  up  of  m  angles,  of  which  n  are 
required  to  make  up  the  whole  circumference  around  P. 

Therefore  the  ratio  of  the  angle  APB  to  a  circumference 

is  m  :  n. 

4.  Because  these  ratios  are  equal  however  great  the  num- 
bers m  and  w,  they  remain  equal  when  the  angle  of  the  lune 
and  the  circumference  are  incommensurable.     Therefore 

Area  of  lune  :  surface  of  sphere  ::  angle  of  lune  :  circumf. 

Q.E.D. 


aUPERFIOIAL  MEASUREMENT.  37^ 

855.  Corollary.  I.  U  there  ai-e  lunes  of  andcs  A    B 
etc.,  we  have  '      ' 

Area (lune  A  +  luuo  J5-f  etc.)  =  area  lune  (^  +  i?  +  etc.). 
of  \^\      ^'''''  ^'    ■^'^'^  ''"''*  ""^  **  liomiBphcre  is  that  of  a  lune 

Theorem  X. 

/•  ^^yV  P^  ^r^  ^-^  ""  spherical  triajigle  is  prcmor- 
tional  to  its  spherical  excess. 

Proof.   Let  .li?(7  be  the  triangle.   Con tmue  any  one  side, 
as  Bty  so  as  to  form  the  great  circle 
BGEF.     Continue  the  other  two  sides    b 
till  they  meet  this  circle  in  E  and  F    /( 
Then— 

1.  Area  ABC -}■  area  ACF  =  lune  '      . 
BABC  =  lune  of  angle  B.  \/ 

Area  ^^C+  area^J5i?^=  luneC^i^^  fV"'"^'^*^ 
=  lune  of  angle  C.  ^^"-., 

Area  ^^C7+area  AFB=  lune  ACA^B  =  Inne  It  Inglo  A. 

2.  BeomseBCBFiB  a  great  circle,  the  sum  of  thi^ffur 
areas  ABC,  A CE,  ABF,  and  ^^^  is  a  hemisphere.  S 
fore,  if  we  add  the  three  equations  (1),  we  have 

2 area ^5C  + hemisphere  =  lune  angle  (^+^+(7).     (8855) 

/QoL       ""'^  *^®  hemisphere  is  the  same  as  a  lune  of  180° 
(§  856),  we  may  write  the  last  equation 

3  area  ABC -\-  lune  180°  =  lune  angle  (^  +  J5  4-  (7V 
whence,  by  transposition,  ^      ■       t"    /> 

2  area  ABC=  lune  {A-\-B-\-C-  180°) 
and  Area  ABC  =  \  lune  (^  +  5  +  c  -  180°)  • 

nv  ^'  ^*t^.+  ^r  ^^^°  ^''  ^y  definition,   the  spherical 
excess  of  the  triangle  ABC    Because  the  area  of  the  lune  is 
proportional  to  its  angle,  the  area  ABCi%  proportional  to  the  ' 
same  angle  or  to  ^  +  j5  +  C  -  180°.     Q.E.D 

Corollary.    If  the  three  vertices  of  a  triangle  are  on  one 
great  circle    its  three  sides  will  coincide  with  that  circle 
and  each  of  its  angles  wiU  be  180°.    Its  area  will  then  be  a 
nemispnere,  and  its  spherical  excess  3.180°  —  180°  =  36"° 


372 


BOOK  XI.    MEASUREMENT  OF  aOLWa. 


I 


Doubling  these  quantities,  wo  have  the  area  of  the  whole 
sphere  corresponding  to  a  spherical  excess  of  720°.     Henco 

868.  The  area  of  a  spherical  triangle  is  to  that  of  the 
whole  sphere  as  its  spherical  excess  is  to  720°. 

Scholium,  Every  spherical  triangle  divides  the  surface  of 
the  sphere  into  two  parts,  of  which  one  may  be  considered 
within  the  triangle,  and  the  other  without  it.  We  may  con- 
sider either  of  these  parts  as  the  area  of  the  triangle,  if,  in 
applying  the  preceding  proposition,  we  measure  the  angles 
through  that  part  of  the  spherical  surface  whose  areas  we  are 
considering.  If  the  inner  angles  are  A,  B,  and  C,  the^anglcs 
measured  on  the  outer  surface  will  be  360°  —  A,  360°  —  B, 
and  360°  -  G  (cf.  §§  25-27).  Subtracting  180°  from  the  sum 
of  these  three  angles  gives  900°  -  (A -^  B -\- C)  as  the  outer 
spherical  excess.  If  the  inner  area  becomes  indefinitely  small, 
AJ^B-\-C  will  approach  to  180°,  and  the  outer  spherical 
excess  will  approach  to  900°  -  180°  =  720°,  which  is  there- 
fore the  spherical  excess  for  the  outer  angles  of  the  triangle 
whose  outer  area  is  that  of  the  whole  sphere. 

869.  Def.  A  zone  is  that  portion  of  the  surface  of 
a  sphere  contained  between  two  parallel  circles  of  the 
sphere. 

860.  Def.  The  altitude  of  a  zone  is  the  perpen- 
dicular distance  between  the  planes  of  its  bounding 
circles. 

Theorem  XL 

861.  The  area  of  a  zone  is  equal  to  the  product 
of  Us  altitude  hy  the  circumference  of  a  great  circle. 

Proof.  Let  HKRSLI  be  a  zone  of  a  sphere  whose  centre 
is  at  0,  and  let  the  plane  of  the  paper  be  a  section  of  the  sphere 
through  the  centre,  perpendicular  to  the  base  HI  of  the  zone. 

The  zone  is  then  generated  by  the  motion  of  the  arc  HKR 
around  the  axis  OT,  perpendicular  to  its  base. 

In  the  arc  HKR  inscribe  the  chords  HK,  KR.  Let  M  be 
the  middle  point  of  HK.  Join  OM,  and  from  M  drop  the 
perpendicular  MD  upon  OT,    Then— 


BVPEUmOIAL  MEASUliEMENT, 


le  whole 
tlenco 

at  of  the 

arface  of 
msiderod 
may  con- 
rle,  if,  in 
le  angles 
^  wo  are 
[le  angles 

50°     -    By 

L  the  sum 
;he  outer 
jly  small, 
spherical 
is  there- 
)  triangle 

irface  of 
3  of  the 


perpen- 
ounding 


873 


product 
it  circle. 
)se  centre 
bhe  sphere 
:  the  zone, 
arc  HKR 

Let  Jf  be 
drop  the 


1.  By  the  revolution  around  the  axis  OT,  the  chord  UK 
will  describe  the  lateral  surface  of  '.he  frustrum  of  a  cone  of 


"^^^l  Sn  *'"®?  ""'"  ^^  ^^  X  circumference  of  a  circle  of 
which  MD  IS  the  radius  (§  848).     That  is, 

Area  of  frustrum  =  ^nMD.HK.  (8  484) 

2.  Because  OMK  is  a  right  angle, 

also  ^^^^  ^^^  ^  ^°°^P*  ^^^  =  ^°°^P-  ^^^y 

Angle  Q  =  angle  D  =  right  angle. 

Therefore  the  triangles  OMD  and  HKO  are  similar,  and 

HE  :  KQ  ::  MO  :  MDi 
whence 

MD.KH=MO.EG, 

3.  Hence,  from  (1), 

Area  of  frustrum  =  2;r.  OM.KO, 

4.  In  the  same  way, 

Area  of  frustrum  KR8L  =  ^nOP  x  alt.  of  frustrum. 

Inscribe  in  the  arc  IfKM  an  indefinite  number  of  equal 
chords  and  consider  the  frustra  they  describe.     Then— 

6.  The  perpendiculars  OM,  OF,  etc.,  will  approach  the 
radius  of  the  sphere  as  their  limit. 

6.  The  sum  of  the  lateral  surfaces  of  all  the  frustra  will 
approach  the  surface  of  the  zone  as  their  limit. 

7.  Because  the  area  of  each  frustrum  approaches  the  limit 

27r  X  radius  of  sphere  X  alt.  of  frustrum, 
the  sum  of  all  the  areas  will  approach  the  limit 

27e  radius  of  sphere  x  sum  of  altitudes  of  frustra 
=  27r  radius  of  sphere  x  alt.  of  zone, 
which  limit  is  the  area  of  the  zone. 


374 


BOOK  XL    MKA8UREMENT  OF  80LID8, 


II 

II 


8.  But  ^n  radius  of  sphere  =  circumf.  of  great  circle. 

Hence: 
Area  of  zone  =  alt.  of  zone  X  circ.  of  groat  circle.    Q.E.D. 

Corollary  1.     Let  AB  and  CD  bo  two  parallel  tangent 

planes  to  a  sphere.    Since  the  preceding  a — ;p^ — ^;^: b 

theorem  is  true  of  zones  of  all  altitudes, 

it  will  remain  true  how  near  soever  we 

suppose  the  bases  of  the  zones  to  the 

tangent  planes.     If,  then,  we  suppose 

the  bases  of  the  zones  to  approach  the 

tangent  planes  as  their  limit,  the  alti-  ^ 

tude  of  the  zone  will  approach  to  the  diameter  of  the  sphere, 

and  its  surface  to  the  surface  of  the  sphere.     Hence: 

862.  The  entire  surface  of  a  sphere  is  equal  to  the  pro- 
duct of  its  diameter  into  its  circumference. 

Cor,  2.  If  we  put  r  for  the  radius  of  the  sphere,  we  have 

Diameter  =  2r, 
Circumference  =  2;rr; 

whence  .  ^     . 

Surface  =  47rr". 

Now  we  have  found  the  area  of  a  circle  of  radius  r  to  be 
Trr*  (§480).     Hence: 

863.  The  area  of  the  surface  of  a  sphere  is  equal  to  the 
area  of  four  great  circles. 

The  area  of  a  hemisphere  is  equal  to  twice  that  of  a  great 

circle. 


•  »  . 


CHAPTER   II. 

VOLUMES  OF  SOLIDS. 


864.  Bef.    The  volume  of  a  solid  is  tlie  measure 
of  its  magnitude. 

865.  Def,  The  base  of  a  solid  is  that  one  of  its 
faces  which  we  select  for  distinction. 

866.  Def,    The  altitude  of  a  solid  is  the  perpen- 


VOLUMEQ, 


875 


ont'bJ^^^  '""^  *^  '^^^^^*  P--^  «P-  the  plaae 
of  S  ^  "^^^  ParaUelopiped  is  a  pamllelopiped 

any  angle  with  each  other,  whereas  the  rectangur.7pard 
lelopiped  has  all  its  faces  perpendicular  to  each  other 

««if  ^?*  ?.^*    '^^^  '^^^^  ®^  volume  is  the  volume  of  a 
cube  of  which  each  edge  is  the  unit  of  length. 

Tolumes  of  Polyhedrons. 

Theorem  XII. 

4H^^\?^^^^  ?^***^*  ^^^^'^^  ^Q.'^l  altitudes  and 
identically  equal  bases  are  identically  equal 

IJ  yiyfo  right  prisms  in  which  ^  J^  J^  i^  U 

and        ^"^  ^^^^^  =  ^*««  A'B'G'D'E'  (identically) 
Alt.  AF  =  alt.  A'F\ 
Conclusion,    The  two  j 

prisms     are     identically     ^/^      _.  ^^.^ 

eqnal.  v<     •'  >r        ..-''7^"^^^!' 

Proof,      Apply    the 
bases  to  each    other   so 
that   they   shall    wholly    , 
coincide.     Let  A'  ~  A,  ^ 
B'  =  B,  etc.    Then—  -  ^  jy    — "c^ 

2.  Because  ^^=  ^'jp' 

q    T     +1,  Point /^=  point  i?*. 

with-a^rJX„"rn::rtr2  r^^**?."*  '^  °-  -U  coincae 

£- — — Q  T.^xtv-i  vx  uiiv  utiier. 


'i" , 


/I/ 


376 


BOOK  XL    MEASUREMENT  OF  SOLIDS. 


4.  Therefore  every  edge  of  the  one  will  coincide  with  a 
corresponding  edge  of  the  other. 

5.  Therefore  every  face  of  the  one  will  coincide  with  the 
jone8pon«ling  face  of  the  other,  whence  the  figures  will  be 
identically  ec^ual.    Q.E.D. 


Theoeem  XIII. 

870.  Bight  prisms  having  equal  bases  and  alti- 
tudes are  equal  in  volume. 

Hypothesis,  JfiVand  PQ,  two  equal  bases  of  right  prisms 
having  equal  altitudes. 


A 

B 

C  / 

/a 

in 

Conclusion.    These  prisms  are  equal  in  volume. 

Proof.  By  the  definition  of  equal  magnitudes  (§13) 
the  hypothesis  implies  that  the  bases  can  be  divided  into  parts 
such  that  each  part  of  the  one  base  shall  be  identically  equal 
to  a  corresponding  part  of  the  other  base. 

Let  A,  B,  and  C  be  the  parts  of  MN,  and  A*,  B\  and 
0'  the  corresponding  identically  equal  parts  of  PQ. 

Let  each  prism  be  divided  into  smaller  prisms  by  planes 
perpendicular  to  the  base,  and  intersecting  it  along  the 
bounding  lines  which  divide  the  bases  into  the  parts  A,  B, 
C,  A',  etc.  Then,  because  the  bases  A.  and  A'  are  identically 
equal  and  have  equal  altitudes, 

Prism  on  base  A  =  prism  on  base  A^  identically.    ( §869) 

In  the  same  way,  each  part  of  the  one  prism  is  identically 
equal  to  a  corresponding  part  of  the  other. 

Therefore  the  two  prisms,  being  made  up  of  these  equal 
part(3,  are  identically  equal.    Q.E.D. 


871.  The  volumes  of  right  prisms  having  equal 
bases  are  proportional  to  their  altitudes. 


VOLUMES. 


377 


e^'S'-  if\ir''r  "''''  ^™-  -  -^-^  f^e 


G 


£ 


8 


Q 


M 


SI 


.>- 


,^-- 


/- 


a 


-1 — 


w 


H 


,^1 


.^'•^■ 


D 


base  ^5CZ>  is  equal  to 
MNOF,  and  the  altitude  AJS 
is  to  the  altitude  MQ  as  m  :  n. 

Conclusion. 
Vol.  Jjy :  vol.  MT  ::m  :  n. 
Proof,  Divide  the  prism 
^iTinto  m  parts  of  equal  alti- 
tude by  planes  parallel  to  the 
base.  Let  prism  MT  be  di- 
vided into  n  parts  in  the  same  way.     Then— 

Tall  Stis^,;"  '^''^  ^^  ^^  -*^  ^  P-^«^  these  parts 

l^laL^mn^r^^'''  P"'"^^  ^"™^  identically  equal 
Dases  and  altitudes,  they  are  equal  in  volume  (8  869) 

3.  Because  the  volume  Jfr  is  composed  of  «  mrts    of 
which  m  parts  make  up  the  volume  ^^,  therefore  ' 

Volume  Aff  :  volume  MT ::  m  :  n.    Q.E  D 

4.  Because  this  is  true  how  great  soever  the  numbers  m 
and  n,  it  remains  true  for  all  cases  (§  359). 

Theorem  XV. 

^./f^^'  ^^  /^^^\  P""^^^^  of  equal  altitudes  the 
mlumes  are  to  each  other  as  the  areas  of  their  bases 

+.  +f '*'''''^'  u^®*  *^®  ^^^^^  ^®  *^  ^^^1^  «ther  as  the  number  m 
to  the  number  ^.  This  means  that  if  the  one  base  be  divTded 
into  m  equal  par  s,  n  of  these  parts  will  make  up  the  other  base 

tuae  to  the  given  prisms. 

These  prisms  will  all  be  equal  (§  870). 

One  given  prism  will  be  made  up  of  m  of  these  eaual 
prisms,  and  the  other  of  n  of  them.  ^ 

Therefore  the  volumes  will  be  to  fiar^h  othov  as  m  to  -z- 

that  is,  tne  volumes  will  be  as  the  areas  of  the  baseJ.    Q.E  d' 

^orollary.    Because  a  parallelepiped  is  a  prism,  we  con^ 


378 


BOOK  XI.    MEASUREMENT  OF  SOLIDS. 


873.  If  a  right  prism  and  a  right  parallelopiped  have 
equal  altitudes,  their  volumes  are  as  the  areas  of  their  bases. 

Theorem  XVI. 

874.  The  volumes  of  rectangular  parallelopipeds 
are  proporticmal  to  the  products  of  their  three  dimen- 
sions. 

Proof  Let  ^  be  a  par- 
allelopiped of  which  the 
dimensions  are  a,  b,  and  h; 
W,  one  of  which  the  dimen- 
sions are  c,  d,  and  k.  * 

Cut  off  from  K  a  paral- 
lelepiped L  of  altitude  k. 
Then— 

1.  Because  L  and  W  have  equal  altitudes, 

Vol.  L  :  vol.  W  ::  area  a.b  :  area  c,d, 
::  ab  :  cd. 

2.  Because  K  and  L  have  equal  bases, 

Vol.  K :  vol.  L  ::  &\t  h  :  alt.  Is. 

3.  Multiplying  these  ratios. 

Vol.  K  :  vol.  W  ::  abh  :  cdk.     Q.E.D. 
Corollary  1.      If   the   dimensions   c,  d,  and  h  of  the 
volume  W  are  each  unity,  the  product  cdJe  will  be  unity,  and 
W  will  be  the  unit  of  volume  (§  868).    The  above  conclusion 
will  then  become     Vol.  K  :  1  ::  abh  :  1, 
which  gives  Vol.  K  =  abh; 

that  is: 

875.  The  volume  of  a  rectangular  parallelopiped  is 
measured  by  the  continued  product 
of  its  three  dimensions. 

Scholium.  If  the  dimensions 
rt,  J,  and  g  are  all  whole  numbers, 
this  result  may  be  shown  in  the 
following  simple  way: 

Being  g  units  in  height,  it 
may  be  divided  up  into  g  layers  each  a  unit  in  height. 


(§872) 
(§416) 

(§871) 


VOLUMES. 


379 


ptd  have 
'bases. 


opipeds 
\dimenr 


w 


U 


(§878) 
(§416) 

(§871) 


h  of  the 
nity,  and 
onclasion 


^piped  is 


b. 


Bemg  h  units  in  breadth,  each  of  these  g  layers  may  be 
divided  into  b  rows,  each  containing  a  units  of  Yolume. 
Thus  the  total  numb.er  of  units  of  volume  will  be  abg. 
Cor,  2.  Because  the  base  of  a  rectangular  parallelepiped 
IS  a  rectangle,  its  area  is  equal  to  the  product  of  its  two 
dimensions.  The  third  dimension  is  then  the  altitude  of  the 
parallelepiped.  The  preceding  result  may  therefore  be  ex- 
pressed in  the  form: 

876.  The  volume  of  a  rectangular  parallelopiped  is  ex- 
pressed by  the  product  of  its  base  into  its  altitude. 

Cor.  3.  Since  a  rectangular  parallelopiped  is  a  kind  of 
right  prism,  every  right  prism  is,  by  §  870,  equal  in  volume  to 
a  rectangular  parallelopiped  having  an  equal  base  and  alti- 
tude.   Therefore  we  conclude : 

877.  The  volume  of  every  right  prism  is  expressed  by  the 
product  of  its  base  and  altitude. 

Theorem  XVII. 

878.  All  parallelopipeds  hamng  the  same  base 
and  equal  altitudes  are  equal  in  xolume. 

Case  I.  Hypothesis.  ABCD-EFQH  ssl^  ABCD-MNOP, 
two    parallelopipeds   hav- 
ing the  same  base  ABCD  P q_  h o 

and  equal  altitudes. 

In  this  case  the  edges    ^ 
FE  and  NM,  also  GH  and 
OP,  are  supposed  to  lie  in 
straight  lines. 

Proof.  1.  Because  EF 
and  MN  are  each  equal 
and  parallel  to  AB^  we  have 

MN  =  EF 
and 

ME  =  NF. 
2.  Considering  the  two  triangular  ^xismsAEM-DHP  and 
BFN-COO,  it  is  proved,  from  the  equality  and  parallelism  of 
all  their  parts,  that  they  are  identically  equal. 


380 


BOOK  XI.    MEASUnSMENT  OF  SOLIDB. 


3.  From  the  solid  ABFM-DCQP  take  away  the  solid 
AEM-DHP,  and  we  have  left  the  parallelopiped  ABOD- 
FFQH, 

4.  From  the  same  solid  ABFM-DGOP  take  away  the 
equal  solid  BFN-CQOy  and  we  have  left  the  parallelopiped 
ABGD-MNOP, 

5.  Therefore 

Volume ABCD-EFGH=  volume  ABCD-MNOP.    Q.E.D. 

Case  II.  Let  the  upper  base  be  in  any  position,  as  IJKL, 

Produce  the  parallel 
edges  FE  and  OH  to  the 
points  M  and  P,  and  pro- 
duce KJ  and  LI  to  N  and 
Mf  forming  the  parallelo- 
gram MNOP.  Join  AM, 
BN,  CO,  DPi  forming 
the  parallelopiped  ABCD- 
MNOP.    Then— 

6.  Because  the  parallelopiped  ABCD-IJKL  has  the  same 
base  and  altitude  as  ABCD-MNOP,  and  has  the  bounding 
edges  JK  and  IL  of  its  upper  face  in  the  same  straight  line 
with  the  edges  PJf  and  ON,  we  have,  by  Case  I., 

Vol.  ABCD-MNOP  =  vol.  ABCD-IJKL, 

7.  We  have,  for  the  same  reason, 

Vol.  ABCD-EFOH  =  vol.  ABCD-MNOP, 

8.  Comparing  with  (6), 

Vol.  ABCD-EFOH  =  vol.  ABCD-IJKL.    Q.E.D. 
Corollary.  Whatever  be  the  oblique  parallelopiped  ABCD- 
IJKL,  we  may  construct  upon  the  same  base  a  right  paral- 
lelopiped ABCD-EFOH  to  which  the  above  demonstration 
will  apply.     Therefore,  from  (§877): 

879.  The  volume  of  any  parallelopiped  is  equal  to  the 
product  of  its  base  and  altitude. 


Theorem  XVIII. 

880.  A  diagonal  plane  divides  any  parallelopiped 
into  two  triangular  prisons  of  equal  volume.  * 


tie  solid 
ABCD- 

way  the 
alopiped 


Q.E.D. 
I IJKL, 


VOLUMES. 


881 


ihe  same 
ounding 
ight  line 


Q.E.D. 
ABGD- 
it  paral- 
istration 

il  to  the 


lopiped 


^  ^^T  b  ^^P^^^''^^-   ABCD-EFOH,  a  right parallelovi- 
ped,  of  which  B DBF  is  a  diagonal  I'^rmmopi 

plane. 

Conclusion. 
Vol.  ABD-EFH=z  vol.  DBG-HFQ. 

Proof.  1.  Because  ABCD  is  a 
parallelogram,  the  diagonal  BD  di- 
vides its  area  into  two  equal  parts. 

2.  Therefore     the     two    right   - 

^riiMtuS^-"'  "^^-^^^  ■"-  ^^--^ "-  -^  t-'^ 

3.  Therefore  these  prisms  are  equal  (§  870).     Q  E  D 

of  wwT  ^I'raF'^^'l'''  ^"^^^^-^FGH,  any  parallelepiped 
of  which  A  CQE  is  a  diagonal  plane. 

Conclusion.     Vol.  .4  CD-EHG  =  vol.  J  CB-EGF. 

Pro(/.^  Through  the  vertices  ^  and  ^  pass  th;  planes 
AUK  and  ^ZifA^  perpendicular  to 
the  parallel  edges  AE,  BF,  CG,  and 

Let  7,  ^,  JT  and  Z,  if,  iV^  be  the 
points  in  which  the  cutting  planes  meet 
these  edges,  produced  when  necessarv. 
Then—  ^ 

1.  Because  the  cutting  planes  are 
perpendicular  to  the  same  edges  AE^  ^k 
etc.,  they  are  parallel.    Therefore  the  ^^"^ 
solid  AUK-ELMN  is  a  right  paral- 
lelepiped, and 

Vol.  AJK  EMN  =  vol.  AJI-EML. 

3.  Because  the  edges  of  both  parallelepipeds  paraUel  to 
AE&ie  also  equal  to  it  (§  687,  cor.  1), 

HN=  ED;     GM  =  CJ;    FL  =  BI-, 

also,  by  comparing  the  sides  of  the  parallelograms, 

EH  =  AD',    EN  =  AE;  EG  =  AC,  etc.; 

and  because  EMG  and  AJO  are  both  right  angles,  by  con- 
struction, "^ 
Angle  EMG  =  angle  AJC, 


(Case  I.) 


382 


BOOK  XL    MEASUREMENT  OF  SOLIDS. 


3.  Therefore  if  the  solid  E-MQHNhe  applied  to  the  solid 
A'JCDK&o  that  the  line  Jg^if  shall  fall  on  AJ,  then 

Triangle  EMN  =  AJK, 
MO  =  JG, 
EO=AG, 
NH=KD, 
QH=CD, 

Therefore  these  two  solids  are  identically  equal. 

4.  If  from  the  prism  ACD-EOH\fQ  take  away  the  solid 
E-MGHN  and  add  the  equal  solid  A-JGDK,  we  shall  have 
the  right  prism  AJK-EMN.     Therefore 

Vol.  ACD-EGH  =  vol.  AJK-EMN. 
6.  In  the  same  way  may  be  shown, 

Vol.  ABG'EFQ  =  vol.  AIJ-ELM. 
6.  Oomparing  with  Case  I., 

Yol  AGD'EGH  =i  yol,  ABG'EFG.    Q.E.D. 


Theokem  XIX. 

881,  The  volume  of  any  prism  is  eqzcaZ  to  the 
prodtcct  of  its  base  by  its  altitude. 

Case  I.    A  triangular  prism. 

Proof.    Let  ABG-DEF  be  any  tri- 
angular prism.     Draw 

BP  II  AG;     GP  II  AB; 
EQ  II  DF;    FQ  \\  DE 
Then— 

1.  Because  ABPG  and  DEQF  are, 
by  construction,  equal  parallelograms 
with  the  sides  of  the  one  parallel  to  the 
coiTesponding  sides  of  the  other,  the 
solid  ABPG'DEQF  ia  a  parallelopiped. 
Therefore 

Vol.  ABPG'DEFQ  =  base  ABPG  X  altitude. 

Area  ABG  =  i  base  ABPG, 


VOLUMES. 


383 


8.  Because  BCFE  is  a  diagonal  plane  of  the  parallelepiped. 

Vol.  ABC'DEF  =  i  vol.  ABPC-DEQFy  (§  880) 

=  i  base  ABPG  X  altitude,  (1) 

=  area  ABC  X  altitude.  (g) 

Case  II.    Any  prism.  Q.^.^. 

Let  ABCDE-FQHIJ  be  any 
prism.  Divide  the  prism  into  tri- 
angular prisms  by  passing  planes 
through  ACHF,  ADIF,  etc.  These 
planes  will  divide  the  bases  into 
triangles.     Then — 

1.  Because  ABC-FGH  is  a  tri- 
angular prism. 

Vol.  ABG-FGH  =  base  ABG 
X  alt.  of  prism. 

2.  In  the  same  way. 

Vol.  AGD-FHI  =  base  AGD  x  alt.  of  prism. 

Vol.  ADE-FIJ  =  base  ABE  x  alt.  of  prism. 

etc.  etc. 

3.  Adding  these  volumes,  we  have 

Sum  of  volumes  =  sum  of  bases  x  alt.  of  prism. 

4.  The  sum  of  these  volumes  makes  up  the  whole  volume 
of  the  prism,  and  the  sum  of  the  triangular  bases  makes  up 
the  whole  base  of , the  prism. 

Therefore  volume  of  prism  =  base  x  altitude.    Q.E.D. 

Theorem  XX. 

882.  All  pyramids  having  equal  bases  and  equal 
altitudes  g,re  equal  in 
liolume. 

Hypothesis.  0-ABCD 
and  P-TUVWZ,  two  pyra- 
mids in  which  area  ABCD 
=  area  r?7FPrZ,  and  alti- 
tude of  0  =  altitude  of  P. 

Gonclusiofi.  The  vol- 
umes of  the  two  pyramids 
are  equal. 


884 


BOOK  XT.    MEASUREMENT  OF  SOLIDS. 


Proof.     Divide  each  pyramid  into  the  same  number  n  of 

slices  by  equidistant  planes  parallel  to  the  base.    Let  us  put 

s,  the  thickness  of  each  slice; 

«,  the  common  altitude  of  each  pyramid; 

b,  the  area  of  the  base  of  each  pyramid. 
Then— 

1.  Because  the  altitude  is  divided  into  n  parts,  the  thick- 
ness of  each  slice  will  be 

n 

2.  Because  the  number  of  slices  is  the  same  in  each  pyra- 
mid, the  distances  of  corresponding  slices  from  the  vertex  will 
be  the  same  in  the  two  pyramids.    If  we  put 

I,  the  distance  of  any  section  from  the  vertex, 

r,  the  area  of  the  section, 
we  shall  have  in  each  pyramid 

r  :  h  ::  r  :  a\  (§  700) 

Also,  the  areas  of  corresponding  sections  are  the  same  in  the 
two  pyramids  (§  701). 

3.  Let  0  and  P  be  two 
corresponding  slices  from  the 
same  pyramid.    Put 

r',  the  area  of  each  upper  base; 

r,  the  area  of  each  lower  base. 
Because  the  lower  base  of  each  is  greater  than  the  upper  base, 
each  slice  is  greater  than  the  prism  of  altitude  s  and  base  r', 
but  less  than  the  prism  of  altitude  s  and  base  r.    Hence 
s  X  r'  <  volume  of  each  slice  <  s  X  r.   . 

4.  Hence  the  difference  between  the  volumes  of  corre- 
sponding slices  of  the  two  pyramids  must  be  less  than  sr  —  «r'; 
that  is,  less  than  s{r  —  r'). 

5.  Let  us  call  the  areas  of  the  several  sections  from  the 
vertex  to  the  base  n,  rg,  rs,  etc.    We  then  have 

Difference  of  top  slices  <  sri. 

Difference  of  second  slices  <  s(r2  —  rj. 

Difference  of  third  slices  <  .?(r.  —  r-). 
etc.  etc. 

Difference  of  bottom  slices  <  s(r„  —  r»_i). 

Adding  up  all  these  differences,  and  noticing  that  r»  =  J,  we 
find 

Difference  of  volumes  of  pyramids  <  sb. 


YOLUMES. 


385 


That  is: 

The  difference  of  the  yolumes  of  the  pyramids  is  less  than 
the  Yolume  of  a  prism  of  equal  base,  and  having  for  its  alti- 
tude the  thickness  of  a  slice. 

6.  But  we  may  take  the  slices  so  thin  that  this  volume  sh 
shall  be  less  than  any  assignable  quantity.  Therefore  the 
volumes  of  the  pyramids  differ  by  less  than  any  assignable 
quantity;  that  is,  they  do  not  differ  at  all.     Q.E.D. 

Theorem  XXI. 

883.  The  wlwme  of  a  pyramid  is  one  third  the 
wlume  of  a  prism  hamng  the  same  base  and  altitude. 

Case  I.     Let  P-ABC  be  a  triangular  pyramid. 

Through  AC  pass  a  plane  AGFD   ^ 

parallel  to  the  opposite  edge  BP,  ^^  '^' 

Complete  the  triangular  prism 
ABC-DP F  by  drawing  the  edges  PD, 
PF,  DF,  AD  and  Ci^paraUel  to  BA, 
BC.ACyBP. 

Divide  the  quadrangular  pyramid 
P-A  GFD  into  two  triangular  prisms 
by  the  plane  PAF,     Then— 

1.  Because  AGFD  is  a  parallelo- 
gram, the  areas  ADF  and  A  GF  are 
equal.    Therefore 

Vol.  P-ADF=  vol.  P-AGF{%  882). 

2.  In  the  same  way,  considering  PBG  and  PFGbs  the  equal 
bases  of  two  triangular  pyramids  having  their  vertices  at  A 

Vol.  A-PBG  =  vol.  A-PFG.  ' 

3.  Comparing  (1)  and  (2),  and  noting 
P-^C^and  A-PFG  9,TQ  the  same  pyra- 
mid, we  see  that  the  prism  ABG-DEF  is 
divided  into  three  equal  pyramids,  of 
which  one  is  the  original  pyramid.    Hence 

Vol.  P'ABG  =  \  vol.  ABG-DEF, 

Case  II.  P-ABGDE,  any  pyramid. 
Through  P  pass  the  planes  PAG, 


386 


BOOK  XI.    MEASUREMENT  OF  SOLIDS. 


FAD,  etc.,  dividing  the  pyramid  into  the  triangular pyramida 
F'ABO,  P-ACD,  etc. 

Let  a  be  the  altitude  of  the  pyramid.    Then,  by  Oase  L, 
Vol.  P-ABG  =  i  prism  ^^C-alt.  a. 
Vol.  P'ACD  =  i  prism  ACD-a\t  a. 
Vol.  P-ADE  =  i  prism  ^i>^-alt.  o. 
The  sum  of  these  pyramids  makes  up  the  given  pyramid, 
and  the  sum  of  the  prisms  is  a  prism  having  the  base 
ABCDB  a,nd  the  altitude  a.    Therefore,  adding, 
Vol.  P-ABGDE  =  i  prism  ABCDE-iXi,  a. 

Q.E.D. 
Corollary.  Because  the  volume  of  a  prism  is  equal  to  the 
product  of  its  base  by  its  altitude,  we  conclude: 

884.  The  volume  of  a  pyramid  is  one  third  the  product 
of  its  base  by  its  altitude. 

Volumes  of  Bound  Bodies. 

Theorem  XXII. 

886.  The  voltime  of  a  cone  is  eqiml  to  one  third 
the  product  of  its  base  hy  its  altitude. 

Proof.    In  the  base  of  the  cone  inscribe  a  regular  poly- 
gon of  any  number  of  sides,  and  upon  it 
erect  a  pyramid  of  which  the  vertex  shall 
be  in  the  vertex  of  the  cone.    Then — 

1.  Because  the  angles  of  the  base  of  the 
pyramid  are  on  the  surface  of  the  cone, 
and  its  vertex  in  the  vertex  of  the  cone,  the 
lateral  edges  of  the  pyramid  will  lie  on  the 
conical  surface,  and  its  altitude  will  be 
equal  to  the  altitude  of  the  cone. 

Let  us  call  a  the  common  altitude  of  cone  and  ppamid. 
Then— 

2.  Volume  of  pyramid  =  ^  «  X  base  of  pyramid. 

3.  Let  the  number  of  sides  of  the  base  of  the  pyramid  be 
indefinitely  increased.  Then  the  base  of  the  pyramid  will 
approach  the  base  of  the  cone  as  its  limit,  and  its  volume  will 
approach  the  volume  of  the  cone  as  its  limit.    Therefore 

Volume  of  cone  =  ^  a  x  base  of  cone.    Q.E.D. 


VOLUMES. 


887 


Theorem  XXIII.  '^ 

886.  The  wlvme  of  a  cylinder  is  eqvM  to  the 
product  qf  its  base  by  its  altitude. 

Proof,  Inscribe  in  the  cylinder  a  prism  of  which  the 
number  of  sides  may  be  increased  without  limit.  Then  the 
base  of  the  prism  will  approach  the  base  of  the  cylinder  as  its 
limit,  and  the  volume  of  the  prism  will  approach  the  volume 
of  the  cylinder  as  its  limit. 

Because  the  volume  of  the  prism  is  continually  equal  to 
the  product  of  its  base  by  its  altitude,  the  volume  of  the 
cylinder  must  also  be  equal  to  the  product  of  its  base  bv  its 
altitude.     Q.E.D. 

Theorem  XXIV. 

887.  The  volume  of  a  sphere  is  equal  to  one  third 
tts  radius  into  the  area  of  its  surface. 

Proof.    Make  a  great  number  of  points  on  the  surface  of 
the  sphere,  and  join  them  by  arcs  of  great 
circles  so  as  to  divide  the  whole  surface 
into  spherical  triangles. 

The  planes  of  these  arcs  will  form  the 
lateral  faces  of  triangular  pyramids  hav- 
ing their  vertices  in  the  centre  of  the 
sphere,  and  the  angles  of  their  bases 
resting  upon  the  surface.  

Because  the  volume  of  each  pyramid  is  ^  base  x  altitude, 
the  combined  volume  of  all  is 

i  sum  of  bases  x  altitude. 

Let  the  number  of  spherical  triangles  be  indefinitely  in- 
creased. Then  the  sum  of  the  bases  of  all  the  pyramids  will 
approach  the  surface  of  the  sphere  as  its  limit,  and  the  alti- 
tudes will  all  approach  the  radius  of  the  sphere  as  their  limit. 
Therefore 

Volume  of  sphere  =  i  radius  X  surface  of  sphere.    Q.E.D. 
Corollary.  We  have  found  (§  862)  for  a  sphere  of  radius  r, 
Surface  of  sphere  =  4n'r". 


888 


BOOK  XI.    MEASUREMENT  OF  SOLIDS. 


Multiplying  this  by  ^r,  we  have 

Volume  of  sphere  =  f^rr*. 

This  result  admits  of  being  memorized  in  the  following 
way.  Suppose  a  cube  circumscribed  about  the  sphere.  Be- 
cause each  of  its  edges  is  2r,  its  volume  will  be  8r'.  Oom- 
paring  with  the  expression  for  the  volume  of  the  sphere,  we 

find 

Vol.  sphere  :  vol.  cube  .:  ^w  :  2. 

Now  if  TT  were  exactly  3,  this  rutio  would  bo  1  :  2;  that  is, 
the  volume  of  the  sphere  \M>ald  be  one  half  that  of  the  cube. 
And  in  reality  the  sphere  is  greater  than  half  the  cube  in  the 
same  ratio  that  n  is  greater  than  3,  which  is  nearly  ^  part 
(§  484). 

Therefore  if  we  fit  a  sphere  into  a  cubical  box,  it  will 
occupy  a  little  more  than  half  the  volume  of  the  box. 


PEOBLEMS  OF  OOMPUTATIOK 


1.  The  altitude  of  a  right  cone  is  4  metres,  and  the  diam- 
eter of  its  base  6  metres.  Compute  its  slant  height,  lateral 
surface,  area  of  base  and  volume. 

Ans.     Slant  height,  6  m. ;  lateral  surface,  — -;  area  of  base, 
9;r;  volume,  IStt. 

2.  The  lateral  aroa  ^  of  a  right  cone  being  given,  what 
relation  must  subsist  between  its  altitude  a  and  the  diameter  D 
of  its  base? 

Ans.     irrD  V{n'  j-  iZ)»)  =  A. 

3.  The  lateral  surface  of  a  right  cone  is  double  the  area  of 
its  base.  What  is  the  ratio  of  its  altitude  to  (he  radius  of  its 
base?  What  must  be  the  ratio  in  order  that  the  lateral  sur- 
face may  be  n  times  the  area  of  the  base? 

Ans.      V'd  and  VnH^, 

4.  Find  the  ratio  of  the  volume  of  a  sphere  to  that  of  its 
right  circumscribed  cylinder. 

Ans.     Vol.  of  sphere  =  f  vol.  of  cylinder. 
Note.    The  circumscribed  cylinder  is  that  whose  bases  and  ele- 
ments are  all  tangents  to  the  sphere. 

6.  The  slant  height  of  a  right  cone  =  diameter  of  its 
base  =  2a.  Express  its  altitude,  lateral  area,  and  volume, 
and  the  radius,  surface,  and  volume  of  its  inscribed  and  cir- 
cumscribed spheres. 

Ans.     Alt.  of  cone,  i^3a;  lateral  area,  2;ra':  volume,  —  : 

Rad.  of  insc.  sphere,  ~;  surface,  —-;  volume,  — '; 

Rad.  of  circ.  sphere,  ~;  surface,  i^;  volume,  ??^ 

V  3  ^  9  4^  * 

6.  The  radius  of  a  sphere  is  bisected  at  right  angles  by  a 
plane.  What  is  the  ratio  of  the  two  parts  into  which  the 
plane  divides  the  spherical  surface? 

Ans.     3  : 1. 

7.  If  a  plane  cut  a  cylinder  at  an  angle  of  45"  with  the 
elements,  what  will  be  the  ratio  of  the  axes  of  the  ellipse  of 
intersection? 

Ans.   i^  :  1. 


THEOEEMS  FOE  EXEEOISE 


IN 


GEOMETRY  OF  THREE  DIMENSIONS. 


BOOK  vm. 


1.  A  line  parallel  to  each  of  two  intersecting  planes  is 
parallel  to  their  line  of  intersection. 

2.  Two  lines,  one  perpendicular  to  one  plane  and  one  to 
another  plane,  form  equal  angles  with  the  planes  to  which 
they  are  not  perpendicular. 

3.  If  a  straight  line  be  perpendicular  to  a  plane,  every  line 
perpendicular  to  that  line  is  parallel  to  the  plane. 

4.  The  supplement  of  any  face  angle  of  a  trihedral  angle 
is  less  than  the  sum,  but  greater  than  the  difference  of  the 
supplements  of  the  two  other  face  angles. 

5.  If,  on  a  line  intersecting  a  plane  perpendicularly,  two 
points,  A  and  B,  equally  distant  from  the  plane  be  taken,  and 
these  points  be  joined  to  three  or  more  points  of  the  plane, 
the  joining  lines  will  form  the  edges  of  two  symmetric  poly- 
hedral angles  having  their  vertices  at  A  and  3, 

6.  If  a  plane  be  passed  through  one  of  the  diagonals  of  a 
parallelogram,  the  perpendiculars  upon  it  from  the  extremi- 
ties of  the  other  diagonal  are  equal. 

7.  If  the  intersections  of  several  planes  are  parallel,  all 
perpendiculars  upon  these  planes  from  the  same  point  in 
space  lie  in  one  plane. 

8.  If  any  number  of  planes  are  respectively  perpendicular 
to  as  many  lines,  and  these  lines  all  lie  in  one  plane,  or  in 
parallel  planes,  the  lines  of  intersection  of  the  planes  are  all 
parallel  to  each  other. 

9.  All  points  whose  projections  upon  a  plane  lie  in  a 
straight  line  are  themselves  in  one  plane.  How  is  this  plane 
defined? 

10.  If  two  straight  lines  are  on  opposite  sides  of  a  plane, 
parallel  to  it,  and  equally  distant  from  it  (but  not  parallel  to 


THEOREMS  IN  GEOMETRY  OF  THREE  DlMENSIOm   391 

each  other),  the  plane  will  bisect  every  line  from  any  point  of 
the  one  line  to  any  point  of  the  other. 

11.  If  any  two  straight  lines  ^  and  5  are  parallel  to  a 
plane  P,  aU  lines  joining  a  point  of  ^  to  a  point  of  B  are  cut 
by  the  plane  P,  internally  or  externally,  into  segments  having 
the  same  ratio. 

12.  Corollary,  If  through  the  ends  of  a  harmonically 
divided  line  two  planes  be  passed  perpendicular  to  the  line, 
and  through  the  harmonic  points  of  division  two  lines  A  and 
B  be  drawn,  each  parallel  to  the  planes,  but  not  in  one  plane, 
then  every  line  joining  a  point  of  ^  to  a  point  of  B  is  cut 
harmonically  by  the  two  planes. 

13.  A  plane  parallel  to  two  sides  of  a  quadrilateral  in 
space  divides  the  other  two  sides  similarly. 


BOOK  IX. 

1.  If  any  two  non-parallel  diagonal  planes  of  a  prism  are 
perpendicular  to  the  base,  the  prism  is  a  right  prism. 

2.  If  the  four  diagonals  of  a  quadrangular  prism  pass 
through  a  point,  the  prism  is  a  parallelopiped. 

3.  A  plane  passing  through  a  triangular  pyramid,  paral- 
lel to  one  side  of  the  base  and  to  the  opposite  lateral  edge, 
mtersects  its  faces  in  a  parallelogram. 

4.  The  four  middle  points  of  two  pairs  of  opposite  edges 
of  a  tnangular  pyramid  are  in  one  plane,  and  at  the  vertices 
of  a  parallelogram. 

Note.  The  six  edges  of  a  triangular  pyramid  may  be  divided  into 
three  pairs,  sach  that  the  two  edges  of  a  pair  do  not  meet  each  other. 
Since  each  edge  meets  two  other  edges  at  one  vertex,  and  two  yet  other 
edges  at  the  adjoining  vertex,  there  is  but  one  edge  left  to  pair  with  it. 
Ihe  pair  is  called  a  pair  of  opposite  edges. 

5.  The  three  lines  joining  the  middle  points  of  the  three 
pairs  of  opposite  edges  of  a  triangular  pyramid  intersect  in  a 
point  which  bisects  them  all. 

6.  The  four  lines  joining  the  vertices  of  a  triangular  pyra- 
mid to  the  centres  of  the  opposite  faces  intersect  in  a  point 
n....^..  -i.Tiviva  ^teuix  ux  iiiiciii  Hi  Liie  raiio  i  :  3. 

Note  The  centre  of  a  triangle  is  the  point  of  intersection  of  its 
three  medial  lines  (§§  168,  169). 


392 


THEOBEMa  FOB  EXERCISE 


7.  The  middle  points  of  the  edges  of  a  regular  tetrahedron 
are  at  the  vertices  of  a  regular  octahedron. 

8.  The  eight  vertices  of  a  cube  are  cut  ofE  by  eight  planes, 
each  passing  through  the  middle  points  of  the  three  edges 
which  diverge  from  each  vertex.  Explain  the  structure  of 
the  polyhedron  thus  formed,  giving  the  number,  form,  and 
relation  of  its  faces,  edges,  and  vertices. 

Describe  its  sympolar  polyhedron,  showing  that  each  face 
is  a  rhombus,  and  explain  the  number  and  form  of  its  edges 
and  vertices. 

Note.  The  sympolar  of  any  polyhedron  may  be  formed  by  draw- 
ing an  edge  across  each  edge  of  the  given  polyhedron,  and  uniting  all 
the  edges  crossing  the  sides  of  each  face  into  a  single  vertex. 


BOOK  X. 


1.  If  lines  be  drawn  from  any  point  of  a  spherical  surface 
to  the  ends  of  a  diameter,  they  will  form  a  right  angle. 

2.  Conversely,  the  locus  of  the  point  from  which  a  finite 
straight  line  subtends  a  right  angle  is  a  spherical  surface  hav- 
ing the  line  for  a  diameter. 

3.  If  any  number  of  lines  in  space  pass  through  a  point, 
the  feet  of  the  perpendiculars  from  another  point  upon  these 
lines  lie  upon  a  spherical  surface. 

4.  If  any  number  of  lines  in  a  plane  pass  through  a  point, 
the  feet  of  the  perpendiculars  upon  these  lines  from  any  point 
not  in  the  plane  lie  on  a  circle. 

6.  If  the  axis  of  an  oblique  circular  cone  is  equal  to  the 
radius  of  the  base,  every  plane  passing  through  the  axis  of  the 
cone  intersects  the  conical  surface  in  lines  forming  a  right 
angle  at  the  vertex.  When  the  axis  of  the  cone  is  less  than 
the  radius  of  the  base,  all  the  angles  thus  formed  are  obtuse, 
and  when  greater  they  are  acute. 

6.  All  parallel  lines  tangent  to  the  same  sphere  intersect 
any  plane  in  an  ellipse. 


t 

JL. 


XJ"' 


BOOK  XI. 

?he  surface  of  a  sphere  is  equal  to  the  lateral  surface  of 
its  circumscribed  cylinder. 
Note.    See  Problem  4,  p.  889. 


IN  OEOMETRT  OF  THREE  DIMENSIOm.    "      393 

3.  If  the  slant  height  of  a  right  cone  is  equal  to  the  diame- 
ter of  its  base,  its  lateral  area  is  double  the  area  of  its  base 

3.  The  lateral  area  of  a  pyramid  is  greater  than  the  area  of 
its  base. 

4.  The  volume  of  a  triangular  prism  is  equal  to  the  area 
of  any  lateral  face  into  half  the  perpendicular  from  the  opno- 
site  edge  upon  that  face.  ^^ 

5.  Any  plane  passing  through  the  middle  points  of  a  pair 
of  opposite  edges  of  a  triangular  pyramid  bisects  its  volume. 

6.  If  the  three  face  angles  of  a  triangular  pyramid  around 
the  vertex  are  all  right  angles,  the  square  of  the  area  of  the 
ba^e  IS  equal  to  the  sum  of  the  squares  of  the  areas  of  the 
tnree  lateral  faces. 

7.  The  bisecting  plane  of  any  edge  angle  of  ft  triangular 
pyramid  divides  the  opposite  edge  into  segments  propor- 
tional to  the  areas  of  the  adjacent  faces. 

8.  Equidistant  parallel  planes  intercept  equal  areas  of  a 
spherical  surface. 


LOCI. 

1.  Find  the  locn  of  the  point  in  space  whose  distances 
from  two  fixed  points  are  in  a  given  ratio. 

3.  Find  the  locus  of  the  point  equally  distant  from  two 
parallel  lines. 

3.  Find  the  locus  of  the  point  equally  distant  from  two 
intersecting  straight  lines. 

4.  Find  the  locus  of  the  point  equally  distant  from  three 
given  points. 

6.  Find  the  locus  of  the  point  equally  distant  from  the 
sides  of  a  triangle. 

6.  Find  the  locus  of  the  point  equally  distant  from  the 
three  edges  of  a  trihedral  angle. 

7.  Two  given  lines  being  on  opposite  sides  of  a  plane 
parallel  to  it,  and  equidistant  from  it,  find  the  locus  of  the 
point  in  the  plane  which  is  equally  distant  from  the  two  lines. 

8.  Find  the  locus  of  the  point  from  whinh  fwn  q/iiQ/»o«+ 
segments  of  the  same  straight  line  subtend  equal  angles. 


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APPEE^DIX. 


NOTES  ON  THE  FUNDAMENTAL  CONCEPTS  OP  GEOMETRY. 


The  true  basis  and  fonii  of  the  fundamental  axioms  and  defini- 
tions  of  Geometry  have  been  the  subject  of  extended  discussion  in 
recent  times,  especially  among  German  mathematicians.  The  follow- 
ing summary  of  conclusions  is  given  partly  to  show  the  direction 
towards  which  these  discussions  tend,  and  partly  to  explain  the 
reasons  for  the  particular  forms  of  definitions  and  axioms  adopted  in 
the  present  work.  Although  the  writer  conceives  that  these  views 
concur  with  the  general  conclusions  of  those  who  have  investigated 
the  subject,  no  one  but  himself  is  to  be  considered  responsible  for  the 
form  in  which  they  are  stated. 

I.  Geometry  has  its  foundation  in  observation.  Clear  conceptions 
of  lines,  as  straight  or  curved;  and,  in  general,  the  idea  of  relative 
positions  in  space,  could  never  be  acquired  except  through  the  eye 
and  touch.  The  ancient  axioms  of  Geometry  proper,  such  as  the 
impossibility  of  two  straight  lines  inclosing  a  space,  the  equality  of 
all  right  angles,  and  the  necessity  of  two  non-parallel  lines  in  the 
same  plane  ultimately  meeting  if  sufficiently  produced,  are  not  to  be 
regarded,  as  they  once  were,  as  necessary  conclusions  apart  from  all 
observation,  but  only  as  necessary  results  of  certain  conceptions 
derived  from  observation.  It  has  in  fact  been  shown  that  a  perfectly 
consistent  Geometry  can  be  constructed  in  which  the  axioms  relatinff 
to  straight  lines  are  not  fulfilled. 

n.  The  general  concepts  of  Geometry— points,  lines,  surfaces,  and 
sohds— are  to  be  regarded  as  attaching  to  material  bodies  rather  than 
as  formed  of  mere  space.  A  geometric  solid,  for  instance,  is  an 
imaginary  material  body  from  which  all  qualities  except  those  of 
extension  and  mobiUty  are  abstracted.  The  quality  of  impenetra- 
bility being  abstracted,  any  two  bodies  may  occupy  the  same  space 
and  may  be  brought  into  absolute  coincidence  if  they  are  identically 
equal  in  their  outlines.  Surfaces,  again,  should  rather  be  considered 
as  extensions  from  which  the  idea  of  thickness  is  abstracted  than  as 
extensions  absolutely  without  thickness.  Similarly,  a  line  need  not 
be  regarded  as  having  no  thickness,  but  may  simply  be  considered  as 


396 


APPENDIX. 


having  the  idea  of  thickness  abstracted.  A  point  is  an  object  the 
magnitude  of  which  we  take  no  account  of. 

This  slight  change  of  conception  may  perhaps  be  regarded  as 
having  little  more  than  metaphysical  interest.  But  it  has  a  certain 
amount  of  practical  value  in  releasing  the  young  mind  from  a  seem- 
ing necessity  of  conceiving  portions  of  pure  space  as  bodies  and 
magnitudes  with  only  one  or  two  dimensions.  In  fact,  it  may  be 
doubted  whether  any  definitions  of  lines,  points,  and  surfaces,  in 
general,  are  of  value  to  a  young  beginner.  He  naturally  falls  into 
the  habit  of  applying  the  terms  the  right  way. 

III.  The  following  considerations  have  led  to  certain  of  the 
primary  definitions  adopted  in  the  present  work. 

1.  It  may  be  doubted  whether  a  straight  line  admits  of  any 
definition  in  the  proper  sense  of  the  term.  A  student  who  does  not 
know  what  a  straight  line  is  before  it  is  defined  will  net  know  in 
consequence  of  the  definition.  The  author  therefore  lays  no  stress 
upon  the  definition  he  has  adopted,  which  is  perhaps  objectionable, 
but  which  has  been  chosen  because  most  readily  understood  by  a 
beginner. 

2.  A  great  majority  of  our  writers  upon  Elementary  Geometry 
make  the  mistake  of  trying  to  include  the  mode  in  which  the  angle 
is  measured  in  the  definition  of  it.  The  system  of  enunciating 
separate  definitions  of  the  angle  and  the  method  of  measuring  it 
has  been  adopted  from  Chauvenet,  and  its  advantages  are  so  obvious 
that  they  need  not  be  pointed  out. 

3.  That  identically  equal  magnitudes  are  those  which  coincide  is 
properly  not  an  axiom,  as  used  in  the  older  geometry,  but  a  defini- 
tion of  the  word  "  equal "  and  its  derivatives.  This  will  be  obvious 
upon  reflecting  that  the  word  must  have  some  definition,  and  that  all 
we  can  mean  by  it  is  that  the  two  objects  to  which  the  term  is 
applied  coincide  when  brought  together,  or  are  made  up  of  coin- 
cident parts.  Had  all  bodies  been  immovable  we  should  never  have 
had  the  idea  of  equality. 

4.  A  statement  of  what  shall  be  meant  by  the  sum  of  two  magni- 
tudes, and  especially  of  two  angles,  is  absolutely  necessary.  The 
want  of  such  a  statement  is  one  of  the  most  serious  defects  in  the 
Geometry  of  Euclid.  Had  Euclid  enunciated  a  general  definition  of 
the  sum  of  two  angles,  and  adhered  to  it,  his  thirteenth  proposition, 
that  the  angles  which  one  straight  line  makes  with  another  upon  one 
side  of  it  are  together  equal  to  two  right  angles,  would  have  been 


unnecessary. 

5.  That  a  straight  line  is  the  shortest  distance  between  i 
of  its  points  is  here  considered  au  axiom  rather  than  a  definition. 


two 


APPENDIX. 


397 


in 


The  r^on  of  placing  it  in  thia  category  is  simply  that  the  idea  of  a 
straight  line  may  be  derived  independently  of  any  compariaoa  of 
general  measures  of  distance  between  the  same  two  points. 

6.  Plane  figures  are  defined  after  the  modern  instead  of  the 
ancient  conceptions.  As  this  will  at  first  sight  strike  the  teacher  of 
the  Euclidean  Geometry  as  one  of  the  most  radical  changes  in  the 
work,  a  comparison  of  the  ideas  on  which  the  two  systems  of  defini- 
tions are  founded  may  be  of  interest. 

The  ancient  geometry  was  primarily  a  science  of  mere  magnitude. 
Solids  were  bodies,  and  plane  figures  were  pieces  of  a  phine  Of 
course  other  conceptions  had  to  be  brought  in,  but  they  were 
regarded  as  subsidiary. 

In  modern  geometry  form  and  position  are  of  equal  importance 
with  magnitude,  and  in  order  that  all  the  conceptions  associated 
with  a  figure  may  come  in  on  terms  of  equality  as  it  were  it  is 
necessary  to  confine  the  definition  of  a  figure  to  what  is  'really 
necessary  to  its  formation.  A  flexibility  and  generaUty  is  thus  given 
to  the  definitions  which  they  cannot  have  under  the  older  form  It 
18  not,  indeed,  claimed  that,  for  the  purpose  of  elementary  instnic 
tion,  one  of  these  systems  of  definitions  has  any  great  advantage  over 
the  other.  But  it  is  important  that  the  definitions  should  accord 
with  the  conceptions  naturally  formed ;  with  the  language  of  every- 
day  life,  and  with  that  of  the  higher  modern  geometry ;  and  theL 
considerations  all  point  to  the  new  system  of  definition.  Let  us  iil^ 
circles  and  polygons  as  examples. 

In  the  older  geometry  a  circle  is  a  round  piece  of  a  plane  or  what, 
in  ordinary  language,  is  called  a  circular  disk.  In  our  laTlangu^^ 
a  circle  IS  a  curved  line  which  the  pupil  can  draw  with  a  pS 

Hne  ZT  ^T^'^y^r^'^  ''  "«^"«  '""'^  '^'^^  -«"« this  curv  d 

lZ^LZ77{r'^  "'  '"'  '^:'''  "^^'^^^"^^  *^«  ^^'^  circumference 
18  applied  to  far  wider  uses.     But  this  nomenclature  is  changed  as 

where  he  finds  the  circle  and  ellipse  treated  as  curves  When  the 
eccentric  ty  vanishes  the  ellipse  is  said  to  become'Tt^hrc'rcum! 
ference  of  a  circle,  but  the  circle  itself.  The  equation  of  the  bound- 
ing curve  of  the  circular  disk  is  also  called  the  equation  of  the 
circle.  In  a  word,  the  old  definition  entirely  vanishes,  Z  a  new 
conception  is  attached  to  the  word. 

A  polygon,  again,  is  completely  determined  by  its  bounding  lines 

IndTed  t  thfvT  '^"'/^^  '"^^^^^  *°y*^-^  ^-t  these  iLes 
Indeed    m  the  higher  modern  geometry  a  Dolvcron  i.  nop«^.,«^ 

XrVn/r"'''"^''  ""^^  ^'"^"^"^  ^'^  ceWn^elations  to'eth 
other,  and  the  more  strongly  a  definition  inconsistent  with  this  ia 


398 


/  PPENDIX. 


impressed  on  the  mind  of  the  beginner,  the  more  difficult  he  findB  It 
to  make  the  necessary  change  in  the  conceptions  attached  to  the  word. 
It  may  seem  that  the  ancient  form  has  some  advantages,  especially  in 
conaldcriug  areas,  and  it  will  also  be  remarked  that  the  word  circum- 
ference is  used  in  the  present  work  in  its  ordinary  acceptation.  A 
glauce  at  the  true  relations  between  geometric  figures  and  words  will 
make  the  state  of  the  case  quite  clear,  and  will  show  a  perfect 
analogy  between  the  system  of  notation  here  adopted  and  the  lan- 
guage of  ordinary  life. 

A  geometric  figure  is  to  be  regarded  as  combining  a  great  number 
of  associated  conceptions,  of  which  a  certain  number  necessarily 
involve  the  others,  and  may  therefore  be  regarded  as  essential.  The 
essential  conceptions  are  those  which  suffice  to  determine  the  figure. 
Since  the  figure  may  be  determined  in  various  ways,  the  only  rule 
that  can  be  followed  is  to  choose  for  a  definition  the  most  simple 
and  easily  understood  set  of  conceptions.  Let  us  consider  a  polygon, 
for  example.  We  have  in  the  polygon  a  collection  of  associated  con- 
ceptions: a  certain  number  of  sides,  a  certain  number  of  interior 
angles,  an  equal  number  of  exterior  angles,  a  form,  an  area,  and  a 
perimeter — the  latter  being  the  sum  of  all  the  sides.  No  one  of  these 
has  any  special  claim  over  the  others  to  be  considered  as  the  measure 
of  the  figure.  Two  different  polygons  equal  in  area  have  no  more 
right  to  be  considered  equal  than  two  other  polygons  of  different 
areas  but  equal  perimeters.  Nor  have  the  concepts  by  which  the 
figure  is  defined,  however  they  may  be  chosen,  any  right  to  be  con- 
sidered as  the  whole  of  the  polygon  because  the  associated  concepts 
equally  belong  to  it.  Hence  the  proper  course  is  to  take  the 
simplest  defining  conception;  namely,  the  lines  which  the  pupil 
draws  when  he  constructs  the  figure,  as  being,  not  necessarily  the 
polygon  itself,  but  the  things  which  determine  or  form  it.  Then  its 
area,  its  angles,  its  perimeter,  its  centre  (if  it  has  one),  its  form,  and 
any  other  concepts  associated  with  it  may  be  separately  considered 
at  pleasure.  Again,  with  the  circle  we  associate  a  circumference,  an 
area,  a  centre,  any  number  of  radii,  and  any  number  of  tangents  we 
choose  to  draw.  The  word  ''circle"  is  properly  applied  only  to  the 
whole  assemblage  of  concepts;  but  since  the  circumscribing  line  is 
the  fundamental  determining  thing,  the  word  can  be  more  properly 
applied  to  it  than  to  any  other  of  the  associated  concepts.  When, 
however,  the  length  of  this  line  comes  into  consideration,  or  when 
the  line  is  to  be  considered  in  antithesis  to  some  other  conception, 
the  centre  for  instance,  then  the  word  circumference  is  used. 

The  accordance  of  this  mode  of  language  with  that  of  ordinary  life 
will  be  seen  by  comparing  it  with  the  ideas  which  we  attach  to  the 


APPENDIX. 


399 


word   ''house."    We  may  equally  define  a  house  as  a  space  sur- 
rounded  by  walls  or  as  walls  enclosing  a  space.    We  habitually  use 
the  word  in  both  senses  without  any  ambiguity  or  confusion.  %^ 
speak  of  building  a  house  when  we  really  mean  building  the  walls 
and  of  living  in  the  house  when  we  mean  living  in  the  inirior         ' 
V.  The  greatest  improvement  in  the  modern  over  the  ancient 
geometry  IS  made  in  the  extension  of  the  idea  of  angular  magS^ 
In  Euclid  ouJy  angles  less  than  ISQo  are  considerc^as  S  anv 
actual  existence.     Angular  measures  equal  to  or  exceed  nrthisHm^t 
are  considered  merely  as  sums  of  angles  to  which  no  viable  gTo 
metric  meaning  is  attached,  and  which  are  in  fact  treated  as  L?!w 
symbolic  entities,  like  the  imaginary  quantities  of  moTe^n   "atCm^^ 
tics.     Some  moderns  have  followed  in  his  footsteps  so  slavishly  L  to 
acually  apprise  the  pupil  that  an  angle  of  180°  i   not  an  angle  Me„t 
the  pupi  might  be  led  into  the  mistake  of  considering  the  sum  o1 
two  right  angles  as  having  some  conceivable  meaning?^ 

We  have  already  mentioned  the  failure  of  Euclid  to  give  any  defi- 
nition  ofthe  sum  of  two  angles.  Without  such  a  definition  we  do  not 
^rZ^  V     Tu  ^  ^^"^  '*°^^''  "•  ^'^^  «"^'^  *  definition  the  sum 

eLln^,-^  /  *"^.r  ^''^°'''  ^^^  ^"^'^  ^^'•'"^d  by  t^«  straight  lines 
extending  from  the  same  vertex  in  opposite  directions. 

In  modern  geometry  angular  measure  is  unlimited,  and  a  civen 
angle  may  have  any  number  of  measures  diflFering  from  each  other  bv 
any  entire  number  of  circumferences.     It  is  not,  however,  advisable 
to  burden  the  beginner  by  attempting  to  impress  this  idea  upon  his 
mind   but  he  should  be  led  up  to  it  gradually.     Hence  in  commenc- 
mg  to  write  the  present  work,  the  author  started  out  by  confining 
angular  measures  to  the  limit  of  180°.    He  soon  found,  however 
that  confusion  would  result  from  attempting  to  keep  within  thi^ 
limit,  especially  m  considering  the  relation  of  angles  inscribed  in  a 
circle      He  therefore  adopted  the  plan  of  extending  angular  meas- 
ures  to  one  circumference,  and  explaining  in  the  beginning  the  two 
measures  of  the  angle.    He  finds  by  expc  rience  that  there  is  no  diffi- 
culty in  making  this  double  measure  clear  to  a  very  young  beginner 


